Answer
$\sqrt 2\pm1$
Work Step by Step
Step 1. Given $\sqrt 2x^2-4x+\sqrt 2=0$, we have $a=\sqrt 2, b=-4, c=\sqrt 2$
Step 2. Use the quadratic formula, we have $x=\frac{4\pm\sqrt {(-4)^2-4(\sqrt 2)(\sqrt 2)}}{2\sqrt 2}=\frac{4\pm\sqrt {8}}{2\sqrt 2}==\frac{2\pm\sqrt {2}}{\sqrt 2}=\sqrt 2\pm1$
