Find the roots of the following quadratic equ | vedclass

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(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 discrimi

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$\frac{1}{2}$ and $-3$

Vedclass · Answer

(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 $-3$.

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $2 y^{2}+5 y-3=0$.

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