Answer
$m=2\pm2\sqrt{3}$
Work Step by Step
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solution/s of the given equation, $
m^2=4m+8
,$ is/are
\begin{array}{l}\require{cancel}
m^2-4m-8=0
\\\\
m=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-8)}}{2(1)}
\\\\
m=\dfrac{4\pm\sqrt{16+32}}{2}
\\\\
m=\dfrac{4\pm\sqrt{48}}{2}
\\\\
m=\dfrac{4\pm\sqrt{16\cdot3}}{2}
\\\\
m=\dfrac{4\pm\sqrt{(4)^2\cdot3}}{2}
\\\\
m=\dfrac{4\pm4\sqrt{3}}{2}
\\\\
m=\dfrac{2(2\pm2\sqrt{3})}{2}
\\\\
m=\dfrac{\cancel{2}(2\pm2\sqrt{3})}{\cancel{2}}
\\\\
m=2\pm2\sqrt{3}
.\end{array}
