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(C) The quadratic formula for the equation $ax^2 + bx + c = 0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Let us check option $(C)$: $\frac{2c}{-b \m

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$\frac{2c}{-b \mp \sqrt{b^2 - 4ac}}$

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(C) The quadratic formula for the equation $ax^2 + bx + c = 0$ is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Let us check option $(C)$: $\frac{2c}{-b \mp \sqrt{b^2 - 4ac}}$.Rationalizing the denominator:$\frac{2c}{-b - \sqrt{b^2 - 4ac}} 	imes \frac{-b + \sqrt{b^2 - 4ac}}{-b + \sqrt{b^2 - 4ac}} = \frac{2c(-b + \sqrt{b^2 - 4ac})}{b^2 - (b^2 - 4ac)} = \frac{2c(-b + \sqrt{b^2 - 4ac})}{4ac} = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$.Similarly,$\frac{2c}{-b + \sqrt{b^2 - 4ac}} = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$.Thus,the expression in option $(C)$ is equivalent to the standard quadratic formula.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

If $ax^2 + bx + c = 0$,then $x =$

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