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