which are the solutions of $x^{2}=-7x - 8$?\n$\\frac{7}{2}-\\frac{\\sqrt{17}}{4},\\frac{7}{2}+\\frac{\\sqrt{1…

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

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

Answer

Answer:

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

Explanation:

Step1: Rewrite the equation

Rewrite $x^{2}=-7x - 8$ as $x^{2}+7x + 8=0$.

Step2: Identify coefficients

For the quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b = 7$, $c = 8$).

Step3: Apply quadratic formula

The quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a = 1$, $b = 7$, $c = 8$ into it: [ \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*} ] So the solutions are $\frac{-7 - \sqrt{17}}{2}$ and $\frac{-7 + \sqrt{17}}{2}$.