Answer
$x_{1}=2+ 2\sqrt{3}$ and $x_{2}=2- 2\sqrt{3}$
Work Step by Step
Given $m^2=4m+8 \longrightarrow m^2-4m-8=0$
$a=1, \ b=-4, \ c=-8$
Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-4)\pm \sqrt{(-4)^2-4\times 1\times (-8)}}{2\times 1} = \dfrac{4\pm \sqrt{16+32}}{2} = \dfrac{4\pm \sqrt{48}}{2} = \dfrac{4\pm \sqrt{4^2\times 3}}{2} = \dfrac{4\pm 4\sqrt{3}}{2} = 2\pm 2\sqrt{3}$
Therefore the solutions are $x_{1}=2+ 2\sqrt{3}$ and $x_{2}=2- 2\sqrt{3}$
