Answer
$\displaystyle x=\frac{-1\pm\sqrt{17}}{2}$
Work Step by Step
We are given:
$x^{2}+x=4$
$x^{2}+x-4=0$
We solve using the quadratic formula ($a=1,\ b=1,\ c=-4$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-(1)\pm\sqrt{(1)^{2}-4*1*-4}}{2*1}$
$\displaystyle x=\frac{-1\pm\sqrt{1+16}}{2}$
$\displaystyle x=\frac{-1\pm\sqrt{17}}{2}$
