which are the solutions of $x^{2}=-7x - 8$?\n$\frac{7}{2}-\frac{sqrt{17}}{4},\frac{7}{2}+\frac{sqrt{17}}{4}$\…

Explanation:

Step1: Rearrange to standard quadratic form

First, rewrite the equation $x^{2}=-7x - 8$ as $x^{2}+7x + 8=0$. The general form of a quadratic equation is $ax^{2}+bx + c = 0$, where in our case $a = 1$, $b=7$, and $c = 8$.

Step2: Apply quadratic formula

The quadratic formula for the solutions of $ax^{2}+bx + c=0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 1$, $b = 7$, and $c = 8$ into the formula: [ \begin{align*} x&=\frac{-7\pm\sqrt{7^{2}-4\times1\times8}}{2\times1}\ &=\frac{-7\pm\sqrt{49 - 32}}{2}\ &=\frac{-7\pm\sqrt{17}}{2} \end{align*} ] The two solutions are $x=\frac{-7-\sqrt{17}}{2}$ and $x=\frac{-7 + \sqrt{17}}{2}$.

Answer:

D. $\frac{-7-\sqrt{17}}{2},\frac{-7+\sqrt{17}}{2}$