Quadratic Equations Questions in English

(C) Given equation: $x^{2}+3x-5=0$.
To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(\frac{3}{2})^{2} = \frac{9}{4}$.
$x^{2}+3x+\frac{9}{4}-5-\frac{9}{4}=0$
$(x+\frac{3}{2})^{2} - \frac{20+9}{4} = 0$
$(x+\frac{3}{2})^{2} - \frac{29}{4} = 0$
$(x+\frac{3}{2})^{2} = (\frac{\sqrt{29}}{2})^{2}$
Taking the square root on both sides:
$x+\frac{3}{2} = \pm \frac{\sqrt{29}}{2}$
$x = -\frac{3}{2} \pm \frac{\sqrt{29}}{2}$
$x = \frac{-3 \pm \sqrt{29}}{2}$
Thus,the solutions are $\frac{-3-\sqrt{29}}{2}$ and $\frac{-3+\sqrt{29}}{2}$.

(D) Given equation: $2x^2 - 7x + 3 = 0$
Divide the entire equation by $2$ to make the coefficient of $x^2$ equal to $1$:
$x^2 - \frac{7}{2}x + \frac{3}{2} = 0$
To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{7}{2})^2 = (\frac{7}{4})^2 = \frac{49}{16}$:
$x^2 - \frac{7}{2}x + \frac{49}{16} - \frac{49}{16} + \frac{3}{2} = 0$
$(x - \frac{7}{4})^2 - \frac{49}{16} + \frac{24}{16} = 0$
$(x - \frac{7}{4})^2 - \frac{25}{16} = 0$
$(x - \frac{7}{4})^2 = (\frac{5}{4})^2$
Taking the square root on both sides:
$x - \frac{7}{4} = \pm \frac{5}{4}$
Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$
Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$
Thus,the solutions are $x = 3$ or $x = \frac{1}{2}$.

(D) Given equation: $4 x^{2}+3 x+5=0$
Divide the entire equation by $4$ to make the coefficient of $x^2$ equal to $1$:
$x^{2} + \frac{3}{4}x + \frac{5}{4} = 0$
To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{3}{4})^2 = (\frac{3}{8})^2 = \frac{9}{64}$:
$x^{2} + \frac{3}{4}x + \frac{9}{64} - \frac{9}{64} + \frac{5}{4} = 0$
$(x + \frac{3}{8})^2 + \frac{-9 + 80}{64} = 0$
$(x + \frac{3}{8})^2 + \frac{71}{64} = 0$
$(x + \frac{3}{8})^2 = -\frac{71}{64}$
Since the square of any real number cannot be negative,there is no real value of $x$ that satisfies this equation. Thus,the equation has no real roots.

(B) Given quadratic equation is $x^{2}+10x+7=0$.
To complete the square,we add and subtract $(\frac{10}{2})^{2} = 25$:
$x^{2}+10x+25-25+7=0$
$(x+5)^{2}-18=0$
$(x+5)^{2} = 18$
Taking the square root on both sides:
$x+5 = \pm\sqrt{18}$
$x+5 = \pm 3\sqrt{2}$
$x = -5 \pm 3\sqrt{2}$
Therefore,the roots are $-5+3\sqrt{2}$ and $-5-3\sqrt{2}$.

(C) To solve $25 x^{2}-30 x+3=0$ by completing the square,we first isolate the constant term: $25 x^{2}-30 x = -3$.
Divide the entire equation by $25$ to make the coefficient of $x^2$ equal to $1$: $x^{2} - \frac{30}{25} x = -\frac{3}{25}$,which simplifies to $x^{2} - \frac{6}{5} x = -\frac{3}{25}$.
To complete the square,add $(\frac{1}{2} \times \text{coefficient of } x)^2 = (\frac{1}{2} \times -\frac{6}{5})^2 = (-\frac{3}{5})^2 = \frac{9}{25}$ to both sides:
$x^{2} - \frac{6}{5} x + \frac{9}{25} = -\frac{3}{25} + \frac{9}{25}$.
This gives $(x - \frac{3}{5})^2 = \frac{6}{25}$.
Taking the square root of both sides: $x - \frac{3}{5} = \pm \frac{\sqrt{6}}{5}$.
Therefore,$x = \frac{3}{5} \pm \frac{\sqrt{6}}{5}$,which results in $x = \frac{3 \pm \sqrt{6}}{5}$.
Thus,the roots are $\frac{3-\sqrt{6}}{5}$ and $\frac{3+\sqrt{6}}{5}$.

(D) To solve the quadratic equation $16 x^{2}-24 x-1=0$ by the method of completing the square,we first rewrite the equation as $16 x^{2}-24 x = 1$.
Divide the entire equation by the coefficient of $x^{2}$,which is $16$:
$x^{2} - \frac{24}{16} x = \frac{1}{16}$
$x^{2} - \frac{3}{2} x = \frac{1}{16}$
Now,add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-\frac{3}{2}$,so half of it is $-\frac{3}{4}$,and its square is $\frac{9}{16}$:
$x^{2} - \frac{3}{2} x + \frac{9}{16} = \frac{1}{16} + \frac{9}{16}$
This simplifies to:
$(x - \frac{3}{4})^{2} = \frac{10}{16}$
Taking the square root of both sides:
$x - \frac{3}{4} = \pm \sqrt{\frac{10}{16}}$
$x - \frac{3}{4} = \pm \frac{\sqrt{10}}{4}$
Solving for $x$:
$x = \frac{3}{4} \pm \frac{\sqrt{10}}{4}$
$x = \frac{3 \pm \sqrt{10}}{4}$
Thus,the roots are $\frac{3+\sqrt{10}}{4}$ and $\frac{3-\sqrt{10}}{4}$.

(A) Given equation: $3 y^{2}+7 y-20=0$
Divide the entire equation by $3$ to make the coefficient of $y^{2}$ equal to $1$:
$y^{2}+\frac{7}{3} y-\frac{20}{3}=0$
To complete the square,add and subtract the square of half the coefficient of $y$,which is $(\frac{1}{2} \times \frac{7}{3})^{2} = (\frac{7}{6})^{2} = \frac{49}{36}$:
$y^{2}+\frac{7}{3} y + \frac{49}{36} - \frac{49}{36} - \frac{20}{3} = 0$
$(y+\frac{7}{6})^{2} - (\frac{49}{36} + \frac{240}{36}) = 0$
$(y+\frac{7}{6})^{2} - \frac{289}{36} = 0$
$(y+\frac{7}{6})^{2} = (\frac{17}{6})^{2}$
Taking the square root on both sides:
$y+\frac{7}{6} = \pm \frac{17}{6}$
Case $1$: $y = \frac{17}{6} - \frac{7}{6} = \frac{10}{6} = \frac{5}{3}$
Case $2$: $y = -\frac{17}{6} - \frac{7}{6} = -\frac{24}{6} = -4$
Thus,the roots are $-4$ and $\frac{5}{3}$.

(B) Given equation: $5x^{2}-4x-10=0$
Divide the entire equation by $5$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2}-\frac{4}{5}x-2=0$
To complete the square,add and subtract the square of half the coefficient of $x$,which is $(\frac{1}{2} \times \frac{4}{5})^{2} = (\frac{2}{5})^{2} = \frac{4}{25}$:
$x^{2}-\frac{4}{5}x + \frac{4}{25} - \frac{4}{25} - 2 = 0$
$(x-\frac{2}{5})^{2} - (\frac{4}{25} + 2) = 0$
$(x-\frac{2}{5})^{2} - (\frac{4+50}{25}) = 0$
$(x-\frac{2}{5})^{2} - \frac{54}{25} = 0$
$(x-\frac{2}{5})^{2} = \frac{54}{25}$
Taking the square root on both sides:
$x-\frac{2}{5} = \pm \sqrt{\frac{54}{25}} = \pm \frac{3\sqrt{6}}{5}$
$x = \frac{2}{5} \pm \frac{3\sqrt{6}}{5}$
$x = \frac{2 \pm 3\sqrt{6}}{5}$
Therefore,the roots are $\frac{2-3\sqrt{6}}{5}$ and $\frac{2+3\sqrt{6}}{5}$.

(D) To solve $x^{2}-8x+12=0$ by completing the square:
$1$. Move the constant term to the side: $x^{2}-8x = -12$
$2$. Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-8$, half of it is $-4$, and its square is $(-4)^{2} = 16$.
$3$. $x^{2}-8x+16 = -12+16$
$4$. Rewrite the left side as a perfect square: $(x-4)^{2} = 4$
$5$. Take the square root of both sides: $x-4 = \pm 2$
$6$. Solve for $x$: $x = 4+2 = 6$ or $x = 4-2 = 2$.
Thus, the roots are $2$ and $6$.

(A) Given the quadratic equation: $x^{2}+2x-2=0$
Step $1$: Move the constant term to the right side: $x^{2}+2x=2$
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $2$,so half of it is $1$,and its square is $1^{2}=1$. Adding $1$ to both sides:
$x^{2}+2x+1=2+1$
Step $3$: Write the left side as a perfect square: $(x+1)^{2}=3$
Step $4$: Take the square root of both sides: $x+1 = \pm\sqrt{3}$
Step $5$: Solve for $x$: $x = -1 \pm \sqrt{3}$
Thus,the roots are $-1+\sqrt{3}$ and $-1-\sqrt{3}$.

(B) Given equation: $x^{2}-4x-8=0$
Step $1$: Shift the constant term to the right side: $x^{2}-4x=8$
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-4$,so half of it is $-2$,and its square is $(-2)^{2}=4$.
$x^{2}-4x+4=8+4$
Step $3$: Write the left side as a perfect square: $(x-2)^{2}=12$
Step $4$: Take the square root on both sides: $x-2 = \pm\sqrt{12}$
$x-2 = \pm 2\sqrt{3}$
Step $5$: Solve for $x$: $x = 2 \pm 2\sqrt{3}$
Thus,the roots are $2+2\sqrt{3}$ and $2-2\sqrt{3}$.

(A) Given equation: $4x^{2} + 20x + 23 = 0$.
Divide the entire equation by $4$: $x^{2} + 5x + \frac{23}{4} = 0$.
Rewrite as: $x^{2} + 5x = -\frac{23}{4}$.
Add the square of half the coefficient of $x$ (which is $(\frac{5}{2})^{2} = \frac{25}{4}$) to both sides: $x^{2} + 5x + \frac{25}{4} = -\frac{23}{4} + \frac{25}{4}$.
This simplifies to: $(x + \frac{5}{2})^{2} = \frac{2}{4} = \frac{1}{2}$.
Taking the square root on both sides: $x + \frac{5}{2} = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$.
Thus,$x = -\frac{5}{2} \pm \frac{\sqrt{2}}{2} = \frac{-5 \pm \sqrt{2}}{2}$.
Therefore,the roots are $\frac{-5+\sqrt{2}}{2}$ and $\frac{-5-\sqrt{2}}{2}$.

(A) Given equation: $4x^{2} + 4bx - (a^{2} - b^{2}) = 0$.
Divide by $4$: $x^{2} + bx - \frac{a^{2}-b^{2}}{4} = 0$.
Add and subtract $(\frac{b}{2})^{2}$ to complete the square: $x^{2} + bx + (\frac{b}{2})^{2} - (\frac{b}{2})^{2} - \frac{a^{2}-b^{2}}{4} = 0$.
$(x + \frac{b}{2})^{2} = \frac{b^{2}}{4} + \frac{a^{2}-b^{2}}{4} = \frac{a^{2}}{4}$.
Taking square root on both sides: $x + \frac{b}{2} = \pm \frac{a}{2}$.
Case $1$: $x = \frac{a-b}{2}$.
Case $2$: $x = \frac{-a-b}{2} = -(\frac{a+b}{2})$.
Thus,the roots are $-\left(\frac{a+b}{2}\right), \left(\frac{a-b}{2}\right)$.

(A) Given equation: $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$.
To complete the square,we add and subtract the square of half the coefficient of $x$,which is $(\frac{\sqrt{3}+1}{2})^{2}$.
$x^{2}-(\sqrt{3}+1) x + (\frac{\sqrt{3}+1}{2})^{2} - (\frac{\sqrt{3}+1}{2})^{2} + \sqrt{3} = 0$.
$(x - \frac{\sqrt{3}+1}{2})^{2} = (\frac{\sqrt{3}+1}{2})^{2} - \sqrt{3}$.
$(x - \frac{\sqrt{3}+1}{2})^{2} = \frac{3+1+2\sqrt{3}}{4} - \sqrt{3} = \frac{4+2\sqrt{3}-4\sqrt{3}}{4} = \frac{4-2\sqrt{3}}{4} = \frac{(\sqrt{3}-1)^{2}}{4}$.
Taking square root on both sides: $x - \frac{\sqrt{3}+1}{2} = \pm \frac{\sqrt{3}-1}{2}$.
Case $1$: $x = \frac{\sqrt{3}+1 + \sqrt{3}-1}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Case $2$: $x = \frac{\sqrt{3}+1 - (\sqrt{3}-1)}{2} = \frac{\sqrt{3}+1-\sqrt{3}+1}{2} = \frac{2}{2} = 1$.
Thus,the roots are $\sqrt{3}$ and $1$.

(B) Given equation: $x^{2}-4 \sqrt{2} x+6=0$.
To complete the square,we rewrite the equation as $x^{2}-2(x)(2 \sqrt{2}) = -6$.
Add $(2 \sqrt{2})^{2} = 8$ to both sides: $x^{2}-2(x)(2 \sqrt{2}) + (2 \sqrt{2})^{2} = -6 + 8$.
This simplifies to $(x-2 \sqrt{2})^{2} = 2$.
Taking the square root on both sides: $x-2 \sqrt{2} = \pm \sqrt{2}$.
Case $1$: $x = 2 \sqrt{2} + \sqrt{2} = 3 \sqrt{2}$.
Case $2$: $x = 2 \sqrt{2} - \sqrt{2} = \sqrt{2}$.
Thus,the roots are $\sqrt{2}$ and $3 \sqrt{2}$.

(C) Given equation: $3x^2 + 11x + 10 = 0$.
Divide the entire equation by $3$ to make the coefficient of $x^2$ equal to $1$: $x^2 + \frac{11}{3}x + \frac{10}{3} = 0$.
Shift the constant term to the right side: $x^2 + \frac{11}{3}x = -\frac{10}{3}$.
Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{11}{3}$,so half of it is $\frac{11}{6}$. The square is $(\frac{11}{6})^2 = \frac{121}{36}$.
$x^2 + \frac{11}{3}x + \frac{121}{36} = -\frac{10}{3} + \frac{121}{36}$.
$(x + \frac{11}{6})^2 = -\frac{120}{36} + \frac{121}{36} = \frac{1}{36}$.
Taking the square root on both sides: $x + \frac{11}{6} = \pm \frac{1}{6}$.
Case $1$: $x + \frac{11}{6} = \frac{1}{6} \implies x = \frac{1}{6} - \frac{11}{6} = -\frac{10}{6} = -\frac{5}{3}$.
Case $2$: $x + \frac{11}{6} = -\frac{1}{6} \implies x = -\frac{1}{6} - \frac{11}{6} = -\frac{12}{6} = -2$.
Thus,the roots are $-\frac{5}{3}$ and $-2$.

(N/A) Given equation: $2x^{2} + x + 4 = 0$.
Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{1}{2}x + 2 = 0$.
Rewrite the equation as $x^{2} + \frac{1}{2}x = -2$.
To complete the square,add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{1}{2}$,so half of it is $\frac{1}{4}$. The square is $(\frac{1}{4})^{2} = \frac{1}{16}$.
$x^{2} + \frac{1}{2}x + \frac{1}{16} = -2 + \frac{1}{16}$.
$(x + \frac{1}{4})^{2} = \frac{-32 + 1}{16} = -\frac{31}{16}$.
Since the square of a real number cannot be negative,there are no real roots for this equation.

(A) Given equation: $2x^{2} - 7x + 3 = 0$.
Divide the entire equation by $2$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} - \frac{7}{2}x + \frac{3}{2} = 0$.
Shift the constant term to the right side:
$x^{2} - \frac{7}{2}x = -\frac{3}{2}$.
Add the square of half the coefficient of $x$ (which is $\frac{1}{2} \times \frac{7}{2} = \frac{7}{4}$) to both sides:
$x^{2} - \frac{7}{2}x + (\frac{7}{4})^{2} = -\frac{3}{2} + (\frac{7}{4})^{2}$.
$(x - \frac{7}{4})^{2} = -\frac{3}{2} + \frac{49}{16}$.
$(x - \frac{7}{4})^{2} = \frac{-24 + 49}{16} = \frac{25}{16}$.
Taking the square root on both sides:
$x - \frac{7}{4} = \pm \frac{5}{4}$.
Case $1$: $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$.
Case $2$: $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$.
Thus,the roots are $3$ and $\frac{1}{2}$.

(B) Given the quadratic equation: $x^{2}+8x+3=0$.
Step $1$: Shift the constant term to the right side: $x^{2}+8x = -3$.
Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $8$,so half of it is $4$,and its square is $4^{2} = 16$.
$x^{2}+8x+16 = -3+16$.
Step $3$: Write the left side as a perfect square: $(x+4)^{2} = 13$.
Step $4$: Take the square root of both sides: $x+4 = \pm\sqrt{13}$.
Step $5$: Solve for $x$: $x = -4 \pm \sqrt{13}$.
Thus,the roots are $-4+\sqrt{13}$ and $-4-\sqrt{13}$.

(D) Given equation: $3x - \frac{2}{x} - 7 = 0$.
Multiply by $x$ to get the standard form: $3x^2 - 7x - 2 = 0$.
Divide by $3$: $x^2 - \frac{7}{3}x - \frac{2}{3} = 0$.
Add and subtract $(\frac{1}{2} \times \text{coefficient of } x)^2 = (\frac{7}{6})^2 = \frac{49}{36}$:
$x^2 - \frac{7}{3}x + \frac{49}{36} - \frac{49}{36} - \frac{2}{3} = 0$.
$(x - \frac{7}{6})^2 = \frac{49}{36} + \frac{24}{36} = \frac{73}{36}$.
Taking square root: $x - \frac{7}{6} = \pm \frac{\sqrt{73}}{6}$.
$x = \frac{7 \pm \sqrt{73}}{6}$.

(A) Given equation: $(2x + 1) - \frac{4}{(2x + 1)} - 3 = 0$.
Let $y = 2x + 1$. Then the equation becomes $y - \frac{4}{y} - 3 = 0$.
Multiplying by $y$ (where $y \neq 0$): $y^2 - 3y - 4 = 0$.
To solve by completing the square: $y^2 - 3y = 4$.
Add $(\frac{3}{2})^2 = \frac{9}{4}$ to both sides: $y^2 - 3y + \frac{9}{4} = 4 + \frac{9}{4}$.
$(y - \frac{3}{2})^2 = \frac{16 + 9}{4} = \frac{25}{4}$.
Taking the square root: $y - \frac{3}{2} = \pm \frac{5}{2}$.
Case $1$: $y = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4$. Since $y = 2x + 1$,$2x + 1 = 4 \implies 2x = 3 \implies x = \frac{3}{2}$.
Case $2$: $y = \frac{3}{2} - \frac{5}{2} = -\frac{2}{2} = -1$. Since $y = 2x + 1$,$2x + 1 = -1 \implies 2x = -2 \implies x = -1$.
The roots are $\frac{3}{2}, -1$.

(B) Given equation: $\frac{2}{x^{2}}-\frac{1}{x}=6$.
Multiply the entire equation by $x^{2}$ (where $x \neq 0$): $2 - x = 6x^{2}$.
Rearrange into standard form: $6x^{2} + x - 2 = 0$.
Divide by $6$: $x^{2} + \frac{1}{6}x - \frac{1}{3} = 0$.
To complete the square,add and subtract $(\frac{1}{2} \times \text{coefficient of } x)^{2} = (\frac{1}{2} \times \frac{1}{6})^{2} = (\frac{1}{12})^{2} = \frac{1}{144}$.
$x^{2} + \frac{1}{6}x + \frac{1}{144} - \frac{1}{144} - \frac{1}{3} = 0$.
$(x + \frac{1}{12})^{2} = \frac{1}{144} + \frac{1}{3} = \frac{1 + 48}{144} = \frac{49}{144}$.
Taking the square root: $x + \frac{1}{12} = \pm \frac{7}{12}$.
Case $1$: $x = \frac{7}{12} - \frac{1}{12} = \frac{6}{12} = \frac{1}{2}$.
Case $2$: $x = -\frac{7}{12} - \frac{1}{12} = -\frac{8}{12} = -\frac{2}{3}$.
Thus,the roots are $-\frac{2}{3}, \frac{1}{2}$.

(C) Given equation: $3x^{2} + 2\sqrt{5}x - 5 = 0$.
Divide the entire equation by $3$ to make the coefficient of $x^{2}$ equal to $1$:
$x^{2} + \frac{2\sqrt{5}}{3}x - \frac{5}{3} = 0$.
Move the constant term to the right side:
$x^{2} + \frac{2\sqrt{5}}{3}x = \frac{5}{3}$.
Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{2\sqrt{5}}{3}$,so half of it is $\frac{\sqrt{5}}{3}$. The square is $(\frac{\sqrt{5}}{3})^{2} = \frac{5}{9}$.
$x^{2} + \frac{2\sqrt{5}}{3}x + \frac{5}{9} = \frac{5}{3} + \frac{5}{9}$.
$(x + \frac{\sqrt{5}}{3})^{2} = \frac{15 + 5}{9} = \frac{20}{9}$.
Taking the square root on both sides:
$x + \frac{\sqrt{5}}{3} = \pm \sqrt{\frac{20}{9}} = \pm \frac{2\sqrt{5}}{3}$.
Case $1$: $x = \frac{2\sqrt{5}}{3} - \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{3}$.
Case $2$: $x = -\frac{2\sqrt{5}}{3} - \frac{\sqrt{5}}{3} = -\frac{3\sqrt{5}}{3} = -\sqrt{5}$.
Thus,the roots are $-\sqrt{5}$ and $\frac{\sqrt{5}}{3}$.

(D) Given equation: $5x(2x - 3) - 4 = 0$
Expand the equation: $10x^2 - 15x - 4 = 0$
Divide by $10$: $x^2 - \frac{15}{10}x - \frac{4}{10} = 0 \Rightarrow x^2 - \frac{3}{2}x - \frac{2}{5} = 0$
Add and subtract $(\frac{1}{2} \times \text{coefficient of } x)^2 = (\frac{1}{2} \times \frac{3}{2})^2 = (\frac{3}{4})^2 = \frac{9}{16}$
$x^2 - \frac{3}{2}x + \frac{9}{16} - \frac{9}{16} - \frac{2}{5} = 0$
$(x - \frac{3}{4})^2 = \frac{9}{16} + \frac{2}{5} = \frac{45 + 32}{80} = \frac{77}{80}$
$x - \frac{3}{4} = \pm \sqrt{\frac{77}{80}} = \pm \frac{\sqrt{77}}{4\sqrt{5}} = \pm \frac{\sqrt{385}}{20}$
$x = \frac{3}{4} \pm \frac{\sqrt{385}}{20} = \frac{15 \pm \sqrt{385}}{20}$
Thus,the roots are $\frac{15+\sqrt{385}}{20}$ and $\frac{15-\sqrt{385}}{20}$.

(A) Given equation: $x + \frac{2}{x} - 8 = 0$.
Multiply the entire equation by $x$ to get: $x^2 - 8x + 2 = 0$.
To solve by completing the square,we rewrite the equation as: $x^2 - 8x = -2$.
Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-8$,so half of it is $-4$,and its square is $(-4)^2 = 16$.
Adding $16$ to both sides: $x^2 - 8x + 16 = -2 + 16$.
This simplifies to: $(x - 4)^2 = 14$.
Taking the square root on both sides: $x - 4 = \pm \sqrt{14}$.
Therefore,$x = 4 \pm \sqrt{14}$.
The roots are $4 + \sqrt{14}$ and $4 - \sqrt{14}$.

(A) Comparing the given equation $x^{2}-5x-1=0$ with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-5, c=-1$.
First,calculate the discriminant $D = b^{2}-4ac$:
$D = (-5)^{2} - 4(1)(-1) = 25 + 4 = 29$.
Since $D > 0$,the equation has two distinct real roots.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values:
$x = \frac{-(-5) \pm \sqrt{29}}{2(1)} = \frac{5 \pm \sqrt{29}}{2}$.
Thus,the roots are $\frac{5+\sqrt{29}}{2}$ and $\frac{5-\sqrt{29}}{2}$.

(B) Comparing the equation $y^{2}+10y+6=0$ with the standard form $ay^{2}+by+c=0$,we get $a=1, b=10, c=6$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (10)^{2} - 4(1)(6) = 100 - 24 = 76$.
Since $D > 0$,the equation has two distinct real roots.
The quadratic formula is $y = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values,we get $y = \frac{-10 \pm \sqrt{76}}{2(1)}$.
Since $\sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}$,we have $y = \frac{-10 \pm 2\sqrt{19}}{2}$.
Dividing by $2$,we get $y = -5 \pm \sqrt{19}$.
Thus,the roots are $-5+\sqrt{19}$ and $-5-\sqrt{19}$.

(A) Comparing the given equation $2 y^{2}+5 y-3=0$ with the standard form $a y^{2}+b y+c=0$,we get $a=2$,$b=5$,and $c=-3$.
First,calculate the discriminant $D = b^{2}-4ac$:
$D = (5)^{2} - 4(2)(-3)$
$D = 25 + 24 = 49$.
Since $D > 0$,the equation has two distinct real roots.
Using the quadratic formula $y = \frac{-b \pm \sqrt{D}}{2a}$:
$y = \frac{-5 \pm \sqrt{49}}{2(2)}$
$y = \frac{-5 \pm 7}{4}$.
Case $1$: $y = \frac{-5+7}{4} = \frac{2}{4} = \frac{1}{2}$.
Case $2$: $y = \frac{-5-7}{4} = \frac{-12}{4} = -3$.
Thus,the roots of the given equation are $\frac{1}{2}$ and $-3$.

(B) Given equation: $(x+4)(x+5)=3(x+1)(x+2)+2x$
Expanding both sides:
$x^2 + 9x + 20 = 3(x^2 + 3x + 2) + 2x$
$x^2 + 9x + 20 = 3x^2 + 9x + 6 + 2x$
Rearranging terms to form $ax^2 + bx + c = 0$:
$x^2 - 3x^2 + 9x - 9x - 2x + 20 - 6 = 0$
$-2x^2 - 2x + 14 = 0$
Dividing by $-2$:
$x^2 + x - 7 = 0$
Comparing with $ax^2 + bx + c = 0$,we get $a=1, b=1, c=-7$.
Discriminant $D = b^2 - 4ac = (1)^2 - 4(1)(-7) = 1 + 28 = 29$.
Since $D > 0$,real and distinct roots exist.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-1 \pm \sqrt{29}}{2(1)}$
$x = \frac{-1 \pm \sqrt{29}}{2}$
Thus,the roots are $\frac{-1+\sqrt{29}}{2}$ and $\frac{-1-\sqrt{29}}{2}$.

(C) Given equation: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}$
Taking $LCM$ on the left side: $\frac{(x+2) + 2(x+1)}{(x+1)(x+2)} = \frac{4}{x+4}$
$\frac{x+2+2x+2}{x^2+3x+2} = \frac{4}{x+4}$
$\frac{3x+4}{x^2+3x+2} = \frac{4}{x+4}$
Cross-multiplying: $(3x+4)(x+4) = 4(x^2+3x+2)$
$3x^2 + 12x + 4x + 16 = 4x^2 + 12x + 8$
$3x^2 + 16x + 16 = 4x^2 + 12x + 8$
Rearranging terms: $x^2 - 4x - 8 = 0$
Comparing with $ax^2 + bx + c = 0$,we get $a=1, b=-4, c=-8$.
Discriminant $D = b^2 - 4ac = (-4)^2 - 4(1)(-8) = 16 + 32 = 48$.
Since $D > 0$,real roots exist.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-(-4) \pm \sqrt{48}}{2(1)} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$.
Thus,the roots are $2(1+\sqrt{3})$ and $2(1-\sqrt{3})$.

(D) Comparing the given equation $x^{2}-8x-21=0$ with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-8, c=-21$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (-8)^{2} - 4(1)(-21) = 64 + 84 = 148$.
Since $D > 0$,the equation has two distinct real roots.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-(-8) \pm \sqrt{148}}{2(1)} = \frac{8 \pm \sqrt{4 \times 37}}{2} = \frac{8 \pm 2\sqrt{37}}{2}$.
Dividing by $2$,we get $x = 4 \pm \sqrt{37}$.
Thus,the roots are $4+\sqrt{37}$ and $4-\sqrt{37}$.

(A) The given quadratic equation is $x^{2} + 6x + 5 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 1$,$b = 6$,and $c = 5$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
First,calculate the discriminant $D = b^{2} - 4ac = (6)^{2} - 4(1)(5) = 36 - 20 = 16$.
Since $D > 0$,real solutions exist.
Substituting the values into the formula: $x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2}$.
For the positive sign: $x = \frac{-6 + 4}{2} = \frac{-2}{2} = -1$.
For the negative sign: $x = \frac{-6 - 4}{2} = \frac{-10}{2} = -5$.
Thus,the solutions are $-5$ and $-1$.

(B) The given quadratic equation is $x^{2}-3x-4=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-3, c=-4$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
First,calculate the discriminant $D = b^{2}-4ac = (-3)^{2} - 4(1)(-4) = 9 + 16 = 25$.
Since $D > 0$,the equation has two distinct real roots.
Substituting the values into the formula:
$x = \frac{-(-3) \pm \sqrt{25}}{2(1)}$
$x = \frac{3 \pm 5}{2}$
Case $1$: $x = \frac{3+5}{2} = \frac{8}{2} = 4$.
Case $2$: $x = \frac{3-5}{2} = \frac{-2}{2} = -1$.
Thus,the solutions are $-1$ and $4$.

(C) The given quadratic equation is $5x^{2} + 8x + 3 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 5$,$b = 8$,and $c = 3$.
The discriminant $D$ is given by $D = b^{2} - 4ac$.
$D = (8)^{2} - 4(5)(3) = 64 - 60 = 4$.
Since $D > 0$,real solutions exist.
The quadratic formula is $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values: $x = \frac{-8 \pm \sqrt{4}}{2(5)} = \frac{-8 \pm 2}{10}$.
Case $1$: $x = \frac{-8 + 2}{10} = \frac{-6}{10} = -\frac{3}{5}$.
Case $2$: $x = \frac{-8 - 2}{10} = \frac{-10}{10} = -1$.
Thus,the solutions are $-\frac{3}{5}$ and $-1$.

(A) Given the quadratic equation: $3x^2 + 5\sqrt{2}x - 2 = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we get $a = 3$,$b = 5\sqrt{2}$,and $c = -2$.
The discriminant $D$ is given by $D = b^2 - 4ac$.
$D = (5\sqrt{2})^2 - 4(3)(-2) = (25 \times 2) + 24 = 50 + 24 = 74$.
Since $D > 0$,the equation has two distinct real roots.
The quadratic formula is $x = \frac{-b \pm \sqrt{D}}{2a}$.
Substituting the values: $x = \frac{-5\sqrt{2} \pm \sqrt{74}}{2(3)} = \frac{-5\sqrt{2} \pm \sqrt{74}}{6}$.
Thus,the solutions are $x = \frac{-5\sqrt{2} + \sqrt{74}}{6}$ and $x = \frac{-5\sqrt{2} - \sqrt{74}}{6}$.

(B) For the quadratic equation $ax^{2}+bx+c=0$,the quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.
Comparing $x^{2}+5x+3=0$ with the standard form,we get $a=1, b=5, c=3$.
The discriminant $D = b^{2}-4ac = (5)^{2} - 4(1)(3) = 25 - 12 = 13$.
Since $D > 0$,real solutions exist.
Substituting the values into the formula: $x = \frac{-5 \pm \sqrt{13}}{2(1)}$.
Thus,the solutions are $x = \frac{-5+\sqrt{13}}{2}$ and $x = \frac{-5-\sqrt{13}}{2}$.

(B) The given quadratic equation is $6x^{2}-5x-1=0$.
Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=6$,$b=-5$,and $c=-1$.
The discriminant $D$ is given by $D = b^{2}-4ac$.
$D = (-5)^{2} - 4(6)(-1) = 25 + 24 = 49$.
Since $D > 0$,the roots are real and distinct.
The quadratic formula is $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-(-5) \pm \sqrt{49}}{2(6)} = \frac{5 \pm 7}{12}$.
Case $1$: $x = \frac{5+7}{12} = \frac{12}{12} = 1$.
Case $2$: $x = \frac{5-7}{12} = \frac{-2}{12} = -\frac{1}{6}$.
Thus,the solutions are $1$ and $-\frac{1}{6}$.

(C) The given quadratic equation is $9x^{2} + 7x - 2 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 9$,$b = 7$,and $c = -2$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
First,calculate the discriminant $D = b^{2} - 4ac = (7)^{2} - 4(9)(-2) = 49 + 72 = 121$.
Now,substitute the values into the formula:
$x = \frac{-7 \pm \sqrt{121}}{2(9)}$
$x = \frac{-7 \pm 11}{18}$
Case $1$: $x = \frac{-7 + 11}{18} = \frac{4}{18} = \frac{2}{9}$.
Case $2$: $x = \frac{-7 - 11}{18} = \frac{-18}{18} = -1$.
Thus,the roots of the equation are $\frac{2}{9}$ and $-1$.

(A) The given quadratic equation is $6x^{2} + x - 2 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 6$,$b = 1$,and $c = -2$.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
First,calculate the discriminant $D = b^{2} - 4ac = (1)^{2} - 4(6)(-2) = 1 + 48 = 49$.
Now,substitute the values into the formula:
$x = \frac{-1 \pm \sqrt{49}}{2(6)}$
$x = \frac{-1 \pm 7}{12}$
Case $1$: $x = \frac{-1 + 7}{12} = \frac{6}{12} = \frac{1}{2}$.
Case $2$: $x = \frac{-1 - 7}{12} = \frac{-8}{12} = -\frac{2}{3}$.
Thus,the roots are $1/2$ and $-2/3$.

(B) For the quadratic equation $ax^{2} + bx + c = 0$,the quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
Given equation: $2x^{2} + 5\sqrt{3}x + 6 = 0$.
Here,$a = 2$,$b = 5\sqrt{3}$,and $c = 6$.
First,calculate the discriminant $D = b^{2} - 4ac = (5\sqrt{3})^{2} - 4(2)(6) = (25 \times 3) - 48 = 75 - 48 = 27$.
Now,apply the formula: $x = \frac{-5\sqrt{3} \pm \sqrt{27}}{2(2)} = \frac{-5\sqrt{3} \pm 3\sqrt{3}}{4}$.
Case $1$: $x = \frac{-5\sqrt{3} + 3\sqrt{3}}{4} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$.
Case $2$: $x = \frac{-5\sqrt{3} - 3\sqrt{3}}{4} = \frac{-8\sqrt{3}}{4} = -2\sqrt{3}$.
Wait,checking the options provided,the correct roots are $-\frac{\sqrt{3}}{2}$ and $-2\sqrt{3}$. Since this matches the calculation,the correct option is $B$.

(A) The given quadratic equation is $p^{2} x^{2} + (p^{2} - q^{2}) x - q^{2} = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = p^{2}$,$b = (p^{2} - q^{2})$,and $c = -q^{2}$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
First,calculate the discriminant $D = b^{2} - 4ac = (p^{2} - q^{2})^{2} - 4(p^{2})(-q^{2})$.
$D = p^{4} - 2p^{2}q^{2} + q^{4} + 4p^{2}q^{2} = p^{4} + 2p^{2}q^{2} + q^{4} = (p^{2} + q^{2})^{2}$.
Now,substitute into the formula: $x = \frac{-(p^{2} - q^{2}) \pm \sqrt{(p^{2} + q^{2})^{2}}}{2p^{2}}$.
$x = \frac{-p^{2} + q^{2} \pm (p^{2} + q^{2})}{2p^{2}}$.
Taking the positive sign: $x = \frac{-p^{2} + q^{2} + p^{2} + q^{2}}{2p^{2}} = \frac{2q^{2}}{2p^{2}} = \frac{q^{2}}{p^{2}}$.
Taking the negative sign: $x = \frac{-p^{2} + q^{2} - p^{2} - q^{2}}{2p^{2}} = \frac{-2p^{2}}{2p^{2}} = -1$.
Thus,the roots are $x = \frac{q^{2}}{p^{2}}$ and $x = -1$.

(A) The given quadratic equation is $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = \sqrt{3}$,$b = 10$,and $c = -8 \sqrt{3}$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
First,calculate the discriminant $D = b^{2} - 4ac = (10)^{2} - 4(\sqrt{3})(-8 \sqrt{3}) = 100 + 32(3) = 100 + 96 = 196$.
Now,substitute the values into the formula:
$x = \frac{-10 \pm \sqrt{196}}{2 \sqrt{3}} = \frac{-10 \pm 14}{2 \sqrt{3}}$.
Taking the positive sign: $x = \frac{-10 + 14}{2 \sqrt{3}} = \frac{4}{2 \sqrt{3}} = \frac{2}{\sqrt{3}}$.
Taking the negative sign: $x = \frac{-10 - 14}{2 \sqrt{3}} = \frac{-24}{2 \sqrt{3}} = \frac{-12}{\sqrt{3}} = -4 \sqrt{3}$.
Thus,the roots are $-4 \sqrt{3}$ and $\frac{2}{\sqrt{3}}$.

(C) Given equation: $\frac{1}{x} - \frac{1}{x-2} = 3$.
Taking $LCM$: $\frac{(x-2) - x}{x(x-2)} = 3$.
$\frac{-2}{x^2 - 2x} = 3$.
$-2 = 3x^2 - 6x$.
$3x^2 - 6x + 2 = 0$.
Comparing with $ax^2 + bx + c = 0$,we get $a = 3, b = -6, c = 2$.
Discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$.
Since $D > 0$,real roots exist.
Using quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-(-6) \pm \sqrt{12}}{2(3)} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}$.

(D) Given equation: $x + \frac{1}{x} = 3$.
Multiply the entire equation by $x$ to clear the denominator: $x^2 + 1 = 3x$.
Rearrange into the standard quadratic form $ax^2 + bx + c = 0$: $x^2 - 3x + 1 = 0$.
Here,$a = 1$,$b = -3$,and $c = 1$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Calculate the discriminant $D = b^2 - 4ac = (-3)^2 - 4(1)(1) = 9 - 4 = 5$.
Since $D > 0$,real roots exist.
Substituting the values into the formula: $x = \frac{-(-3) \pm \sqrt{5}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$.
Thus,the solutions are $x = \frac{3+\sqrt{5}}{2}$ and $x = \frac{3-\sqrt{5}}{2}$.

(A) Given the quadratic equation: $3x^{2} + 2\sqrt{5}x - 5 = 0$.
Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 3$,$b = 2\sqrt{5}$,and $c = -5$.
The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.
First,calculate the discriminant $D = b^{2} - 4ac = (2\sqrt{5})^{2} - 4(3)(-5) = 20 + 60 = 80$.
Now,substitute the values into the formula:
$x = \frac{-2\sqrt{5} \pm \sqrt{80}}{2(3)} = \frac{-2\sqrt{5} \pm 4\sqrt{5}}{6}$.
Case $1$: $x = \frac{-2\sqrt{5} + 4\sqrt{5}}{6} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.
Case $2$: $x = \frac{-2\sqrt{5} - 4\sqrt{5}}{6} = \frac{-6\sqrt{5}}{6} = -\sqrt{5}$.
Thus,the solutions are $-\sqrt{5}$ and $\frac{\sqrt{5}}{3}$.

(C) Given equation: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4}$.
Taking $LCM$ on the left side: $\frac{(x+2) + 2(x+1)}{(x+1)(x+2)} = \frac{4}{x+4}$.
Simplify the numerator: $\frac{x+2+2x+2}{x^2+3x+2} = \frac{4}{x+4} \implies \frac{3x+4}{x^2+3x+2} = \frac{4}{x+4}$.
Cross-multiply: $(3x+4)(x+4) = 4(x^2+3x+2)$.
Expand both sides: $3x^2 + 12x + 4x + 16 = 4x^2 + 12x + 8$.
Rearrange to form a standard quadratic equation $ax^2+bx+c=0$: $x^2 - 4x - 8 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-4, c=-8$.
$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-8)}}{2(1)} = \frac{4 \pm \sqrt{16 + 32}}{2} = \frac{4 \pm \sqrt{48}}{2}$.
Since $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$,we get $x = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$.
Thus,the solutions are $2+2\sqrt{3}$ and $2-2\sqrt{3}$.

(D) The given quadratic equation is $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.
Comparing this with the standard form $ax^2 + bx + c = 0$,we get $a = \sqrt{2}$,$b = 7$,and $c = 5\sqrt{2}$.
The discriminant $D$ is given by $D = b^2 - 4ac$.
$D = (7)^2 - 4(\sqrt{2})(5\sqrt{2}) = 49 - 4(5)(2) = 49 - 40 = 9$.
Since $D > 0$,the equation has two distinct real roots.
The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.
$x = \frac{-7 \pm \sqrt{9}}{2\sqrt{2}} = \frac{-7 \pm 3}{2\sqrt{2}}$.
Case $1$: $x = \frac{-7 + 3}{2\sqrt{2}} = \frac{-4}{2\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$.
Case $2$: $x = \frac{-7 - 3}{2\sqrt{2}} = \frac{-10}{2\sqrt{2}} = -\frac{5}{\sqrt{2}}$.
Thus,the solutions are $-\sqrt{2}$ and $-\frac{5}{\sqrt{2}}$.

(C) Let the breadth of the rectangle be $x\,cm$.
Then,by the given condition,the length of the rectangle is $(3x - 2)\,cm$.
Now,by the data,the area of the rectangle is $280\,cm^2$.
$\therefore$ Area of rectangle $= 280$
$\therefore$ Length $\times$ Breadth $= 280$
$\therefore (3x - 2)(x) = 280$
$\therefore 3x^2 - 2x = 280$
$\therefore 3x^2 - 2x - 280 = 0$
$\therefore (3x + 28)(x - 10) = 0$
$\therefore 3x + 28 = 0$ or $x - 10 = 0$
$\therefore x = -\frac{28}{3}$ or $x = 10$
As $x$ is the breadth of a rectangle,it cannot be negative.
$\therefore x = -\frac{28}{3}$ is not possible.
$\therefore x = 10$
Now,the length of the rectangle $= 3x - 2 = 3(10) - 2 = 28\,cm$.
Hence,the length of the given rectangle is $28\,cm$.

(A) Let the smaller of two consecutive even positive integers be $x$.
Then,the greater even positive integer is $x+2$.
According to the problem,the sum of their squares is $244$.
$\therefore x^{2} + (x+2)^{2} = 244$
$\therefore x^{2} + x^{2} + 4x + 4 = 244$
$\therefore 2x^{2} + 4x + 4 - 244 = 0$
$\therefore 2x^{2} + 4x - 240 = 0$
Dividing the equation by $2$,we get:
$\therefore x^{2} + 2x - 120 = 0$
Factoring the quadratic equation:
$\therefore (x+12)(x-10) = 0$
$\therefore x+12 = 0$ or $x-10 = 0$
$\therefore x = -12$ or $x = 10$
Since the required numbers are positive integers,$x = -12$ is not possible.
Therefore,$x = 10$.
The smaller number is $10$ and the greater number is $10+2 = 12$.
Hence,the required consecutive even positive integers are $10$ and $12$.

(C) Let the digit at the ten's place be $x$.
Since the product of the digits is $6$,the digit at the unit's place is $\frac{6}{x}$.
The original number is $10x + \frac{6}{x}$.
When the digits are interchanged,the new number is $10(\frac{6}{x}) + x = \frac{60}{x} + x$.
According to the problem,subtracting $9$ from the original number gives the new number:
$10x + \frac{6}{x} - 9 = \frac{60}{x} + x$
Multiply the entire equation by $x$:
$10x^2 + 6 - 9x = 60 + x^2$
$9x^2 - 9x - 54 = 0$
Divide by $9$:
$x^2 - x - 6 = 0$
$(x - 3)(x + 2) = 0$
Since a digit cannot be negative,$x = 3$.
The original number is $10(3) + \frac{6}{3} = 30 + 2 = 32$.