Answer
$x=\dfrac{2\pm\sqrt{2}}{2}$
Work Step by Step
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions of the given quadratic equation, $
2x^2-4x+1=0
,$ are
\begin{array}{l}\require{cancel}
x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(2)(1)}}{2(2)}
\\\\
x=\dfrac{4\pm\sqrt{16-8}}{4}
\\\\
x=\dfrac{4\pm\sqrt{8}}{4}
\\\\
x=\dfrac{4\pm\sqrt{4\cdot2}}{4}
\\\\
x=\dfrac{4\pm\sqrt{(2)^2\cdot2}}{4}
\\\\
x=\dfrac{4\pm2\sqrt{2}}{4}
\\\\
x=\dfrac{2(2\pm\sqrt{2})}{4}
\\\\
x=\dfrac{\cancel{2}(2\pm\sqrt{2})}{\cancel{2}(2)}
\\\\
x=\dfrac{2\pm\sqrt{2}}{2}
.\end{array}
