Answer
$x_{1}\approx-0.93$
$x_{2}\approx0.93$
Work Step by Step
The quadratic formula will be used, but before doing so substitution will be used:$x^2=u$ therefore:
$u^2+\sqrt2u-2=0$
$u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where a=1, b=$\sqrt2$, and c=-2
$u=\frac{-(\sqrt2)\pm\sqrt{(\sqrt2)^2-4(1)(-2)}}{2(1)}$
$u=\frac{-\sqrt2\pm\sqrt{2+8}}{2}$
$u=\frac{-\sqrt2\pm\sqrt{10}}{2}$
There are two parts:
First:
$x^2=\frac{-\sqrt2+\sqrt{10}}{2}$
$\sqrt{x^2}=\sqrt{\frac{-\sqrt2+\sqrt{10}}{2}}$
$x\approx\pm0.93$
Second:
$x^2=\frac{-\sqrt2-\sqrt{10}}{2}$
$\sqrt{x^2}=\sqrt{\frac{-\sqrt2-\sqrt{10}}{2}}$
It is not possible to solve since there is a negative value inside the radical.
Plugging the first values into the original equation confirms that they are the only real solutions.