Answer
$x_{1}=\dfrac{2 + \sqrt{2}}{2}$ and $x_{2}=\dfrac{2 - \sqrt{2}}{2}$
Work Step by Step
Given $2x^2-4x+1=0 \\ $
$a=2, \ b=-4, \ c=1 \\ $
Therefore, using the quadratic formula: $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} $ we have:
$\dfrac{-(-4) \pm \sqrt{(-4)^2-4 \times 2 \times 1}}{2 \times 2} = \dfrac{4 \pm \sqrt{16-8}}{4} = \dfrac{4 \pm \sqrt{8}}{4} = \dfrac{4 \pm \sqrt{2^2 \times 2}}{4} = \dfrac{4 \pm 2\sqrt{2}}{4} = \dfrac{2 \pm \sqrt{2}}{2} \\ $
Therefore the solutions are $x_{1}=\dfrac{2 + \sqrt{2}}{2}$ and $x_{2}=\dfrac{2 - \sqrt{2}}{2}$