Answer
$m_{1} = 2+\sqrt{2}$ and $m_{2} = 2-\sqrt{2}$
Work Step by Step
Given $\dfrac{1}{4}m^2-m+\dfrac{1}{2} = 0$
$a = \dfrac{1}{4},\ b=-1,\ c =\dfrac{1}{2} $
Using the quadratic formula $\dfrac{-b\pm\sqrt{b^2-4ab}}{2a} , $ we have:
$\dfrac{-(-1)\pm \sqrt{(-1)^2-4\times \dfrac{1}{4}\times \dfrac{1}{2}}}{\dfrac{1}{4}\times 2} = \dfrac{1\pm \sqrt{1-\dfrac{1}{2}}}{\dfrac{1}{2}} = \dfrac{1\pm \sqrt{\dfrac{1}{2}}}{\dfrac{1}{2}} = 2\pm2\sqrt{\dfrac{1}{2}} = 2\pm2\dfrac{1}{\sqrt{2}} = 2\pm\sqrt{2}$
Therefore the solutions are $m_{1} = 2+\sqrt{2}$ and $m_{2} = 2-\sqrt{2}$
