Question

What are the solutions of x^2 = -7x-8

Answer 1

$Solution, solve\:for\:x,\:x^2=-7x-8\quad :\quad x=\frac{-7+\sqrt{17}}{2},\:x=\frac{-7-\sqrt{17}}{2}$

$Steps:$

$x^2=-7x-8$

$\mathrm{Add\:}8\mathrm{\:to\:both\:sides},\\x^2+8=-7x-8+8$

$\mathrm{Simplify},\\x^2+8=-7x$

$\mathrm{Add\:}7x\mathrm{\:to\:both\:sides},\\x^2+8+7x=-7x+7x$

$\mathrm{Simplify},\\x^2+7x+8=0$

$Solve\:with\:the\:quadratic\:formula,\\\mathrm{For\:}\quad a=1,\:b=7,\:c=8:\quad x_{1,\:2}=\frac{-7\pm \sqrt{7^2-4\cdot \:1\cdot \:8}}{2\cdot \:1},\\x=\frac{-7+\sqrt{7^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}:\quad \frac{-7+\sqrt{17}}{2},\\x=\frac{-7-\sqrt{7^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}:\quad \frac{-7-\sqrt{17}}{2}$

$\mathrm{The\:final\:solutions\:to\:the\:quadratic\:equation\:are:},\\x=\frac{-7+\sqrt{17}}{2},\:x=\frac{-7-\sqrt{17}}{2}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$