#### Answer

$x=2\pm\sqrt{2}$

#### Work Step by Step

$x^{2}-4x+2=0$
Let's take the independent term to the right side of the equation:
$x^{2}-4x=-2$
Let's remember, that in order to complete the square, we need to add $(\dfrac{b}{2})^{2}$ to both sides of the equation. $b$ is the coefficient of the first degree term. In this case $b=-4$
$x^{2}-4x+(\dfrac{-4}{2})^{2}=-2+(\dfrac{-4}{2})^{2}$
$x^{2}-4x+4=-2+4$
$x^{2}-4x+4=2$
We have a perfect square trinomial on the left side of the equation. We factor it and the equation becomes:
$(x-2)^{2}=2$
Take the square root of both sides of the equation:
$\sqrt{(x-2)^{2}}=\sqrt{2}$
$x-2=\pm\sqrt{2}$
Solve for $x$:
$x=2\pm\sqrt{2}$