Answer
$\displaystyle x=\frac{-\sqrt{2}\pm\sqrt{10}}{2}$
$x=0.87$ or $x=-2.29$
Work Step by Step
We are given:
$x^{2}+\sqrt{2}x-2=0$
We solve using the quadratic formula ($a=1,\ b=\sqrt{2},\ c=-2$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{-\sqrt{2}\pm\sqrt{(\sqrt{2})^{2}-4*1*-2}}{2*1}$
$\displaystyle x=\frac{-\sqrt{2}\pm\sqrt{2+8}}{2}$
$\displaystyle x=\frac{-\sqrt{2}\pm\sqrt{10}}{2}$
$x=0.87$ or $x=-2.29$
