Answer: [tex]\bold{x=\dfrac{-1+ \sqrt{33}}{2}\qquad and\qquad x=\dfrac{-1- \sqrt{33}}{2}}[/tex]
Step-by-step explanation:
[tex]x^2+x=8\\\\\text{subtract 8 from both sides:}\\x^2+x-8=0\qquad \rightarrow \qquad a=1, b=1, c=-8\\\\\\\text{Use the quadratic formula to solve for x:}\\x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x = \dfrac{-(1)\pm \sqrt{(1)^2-4(1)(-8)}}{2(1)}\\\\\\.\ =\dfrac{-1\pm \sqrt{1+32}}{2}\\\\\\.\ =\dfrac{-1\pm \sqrt{33}}{2}\\\\\\x=\dfrac{-1+ \sqrt{33}}{2}\qquad and\qquad x=\dfrac{-1- \sqrt{33}}{2}[/tex]
