[tex]x^2-4x+2=0 \\ \\
a=1 \\ b=-4 \\ c=2 \\ b^2-4ac=(-4)^2-4 \times 1 \times 2=16-8=8 \\ \\
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4) \pm \sqrt{8}}{2 \times 1}=\frac{4 \pm \sqrt{4 \times 2}}{2}=\frac{4 \pm 2\sqrt{2}}{2}=\frac{2(2 \pm \sqrt{2})}{2}=2 \pm \sqrt{2} \\
\boxed{x=2-\sqrt{2} \hbox{ or } x=2+\sqrt{2}}[/tex]
