#### Answer

$x=\left\{ \sqrt{2}-1,\sqrt{2}+1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\sqrt{2}x^2-4x+\sqrt{2}=0
,$ use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
In the equation above, $a=
\sqrt{2}
,$ $b=
-4
,$ and $c=
\sqrt{2}
.$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{array}{l}\require{cancel}
x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(\sqrt{2})(\sqrt{2})}}{2(\sqrt{2})}
\\\\
x=\dfrac{4\pm\sqrt{16-8}}{2(\sqrt{2})}
\\\\
x=\dfrac{4\pm\sqrt{8}}{2(\sqrt{2})}
\\\\
x=\dfrac{4\pm\sqrt{4\cdot2}}{2(\sqrt{2})}
\\\\
x=\dfrac{4\pm\sqrt{(2)^2\cdot2}}{2(\sqrt{2})}
\\\\
x=\dfrac{4\pm2\sqrt{2}}{2(\sqrt{2})}
\\\\
x=\dfrac{\cancel2^2\pm\cancel2^1\sqrt{2}}{\cancel2^1(\sqrt{2})}
\\\\
x=\dfrac{2\pm\sqrt{2}}{\sqrt{2}}
.\end{array}
Rationalizing the denominator results to
\begin{array}{l}\require{cancel}
x=\dfrac{2\pm\sqrt{2}}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}
\\\\
x=\dfrac{2\sqrt{2}\pm(\sqrt{2})^2}{(\sqrt{2})^2}
\\\\
x=\dfrac{2\sqrt{2}\pm2}{2}
\\\\
x=\dfrac{\cancel2^1\sqrt{2}\pm\cancel2^1}{\cancel2^1}
\\\\
x=\sqrt{2}\pm1
.\end{array}
The solutions are $
x=\left\{ \sqrt{2}-1,\sqrt{2}+1 \right\}
.$