Answer
$m=2\pm\sqrt{2}$
Work Step by Step
The values of $m$ that satisfy the equation $
\dfrac{1}{4}m^2-m+\dfrac{1}{2}=0
$ are
\begin{array}{l}
4\left( \dfrac{1}{4}m^2-m+\dfrac{1}{2}\right)=(0)4
\\\\
m^2-4m+2=0
\\\\
m=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
\\\\
m=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}
\\\\
m=\dfrac{4\pm\sqrt{8}}{2}
\\\\
m=\dfrac{4\pm2\sqrt{2}}{2}
\\\\
m=2\pm\sqrt{2}
\end{array}
