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(A) Given equation is $5x^{2}-2x-10=0$.On comparing with the standard form $ax^{2}+bx+c=0$,we get:$a=5, b=-2, c=-10$.First,we calculate the discrimina

Vedclass · Accepted Answer

$\frac{1+\sqrt{51}}{5}, \frac{1-\sqrt{51}}{5}$

Vedclass · Answer

(A) Given equation is $5x^{2}-2x-10=0$.On comparing with the standard form $ax^{2}+bx+c=0$,we get:$a=5, b=-2, c=-10$.First,we calculate the discriminant $D = b^{2}-4ac$:$D = (-2)^{2} - 4(5)(-10)$$D = 4 + 200 = 204$.Since $D > 0$,the equation has two distinct real roots.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:$x = \frac{-(-2) \pm \sqrt{204}}{2(5)}$$x = \frac{2 \pm \sqrt{4 	imes 51}}{10}$$x = \frac{2 \pm 2\sqrt{51}}{10}$$x = \frac{1 \pm \sqrt{51}}{5}$.Thus,the roots are $\frac{1+\sqrt{51}}{5}$ and $\frac{1-\sqrt{51}}{5}$.

Find whether the following equation has real roots. If real roots exist,find them.
$5x^{2}-2x-10=0$

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Find whether the following equation has real roots. If real roots exist,find them.$5x^{2}-2x-10=0$