Answer
$x_{1}\approx11.66$
$x_{2}\approx0.34$
Work Step by Step
The quadratic formula will be used, but before doing so substitution will be used:$x^{1/2}=u$ therefore:
$u^2-4u+2=0$
$u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where a=1, b=-4, and c=2
$u=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$
$u=\frac{4\pm\sqrt{16-8}}{2}$
$u=\frac{4\pm\sqrt{8}}{2}$
$u=\frac{4\pm\sqrt{2^3}}{2}$
$u=\frac{4\pm2\sqrt{2}}{2}$
$u=\frac{2(2\pm\sqrt{2})}{2}$
$u=2\pm\sqrt{2}$
There are two solutions:
First:
$x^{1/2}=2+\sqrt2$
$(x^{1/2})^2=(2+\sqrt2)^2$
$x\approx11.66$
Second:
$x^{1/2}=2-\sqrt2$
$(x^{1/2})^2=(2-\sqrt2)^2$
$x\approx0.34$
Plugging these values into the original equation does confirm that they are real solutions.