Answer
$x=2\pm\sqrt{3}$
Work Step by Step
Using the properties of equality, the given quadratic equation, $
\dfrac{1}{2}x^2-2x+\dfrac{1}{2}=0
,$ is equivalent to
\begin{array}{l}\require{cancel}
2\left( \dfrac{1}{2}x^2-2x+\dfrac{1}{2} \right)=(0)2
\\\\
x^2-4x+1=0
.\end{array}
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions of the quadratic equation above are
\begin{array}{l}\require{cancel}
x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)}
\\\\
x=\dfrac{4\pm\sqrt{16-4}}{2}
\\\\
x=\dfrac{4\pm\sqrt{12}}{2}
\\\\
x=\dfrac{4\pm\sqrt{4\cdot3}}{2}
\\\\
x=\dfrac{4\pm\sqrt{(2)^2\cdot3}}{2}
\\\\
x=\dfrac{4\pm2\sqrt{3}}{2}
\\\\
x=\dfrac{2(2\pm\sqrt{3})}{2}
\\\\
x=\dfrac{\cancel{2}(2\pm\sqrt{3})}{\cancel{2}}
\\\\
x=2\pm\sqrt{3}
.\end{array}
