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(C) Given equation is $-x^{2}+7x-10=0$.On comparing with the standard form $ax^{2}+bx+c=0$,we get:$a = -1, b = 7, c = -10$.The quadratic formula is gi

Vedclass · Accepted Answer

$2, 5$

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(C) Given equation is $-x^{2}+7x-10=0$.On comparing with the standard form $ax^{2}+bx+c=0$,we get:$a = -1, b = 7, c = -10$.The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.Substituting the values:$x = \frac{-(7) \pm \sqrt{(7)^{2}-4(-1)(-10)}}{2(-1)}$$x = \frac{-7 \pm \sqrt{49-40}}{-2}$$x = \frac{-7 \pm \sqrt{9}}{-2}$$x = \frac{-7 \pm 3}{-2}$.Calculating the two roots:$x_1 = \frac{-7 + 3}{-2} = \frac{-4}{-2} = 2$$x_2 = \frac{-7 - 3}{-2} = \frac{-10}{-2} = 5$.Thus,the roots of the equation are $2$ and $5$.

Find the roots of the quadratic equation by using the quadratic formula: $-x^{2}+7x-10=0$.

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