Answer
$\left\{ 2-2\sqrt{3},2+2\sqrt{3} \right\}$
Work Step by Step
The standard form of the given equation, $
y^2-8=4y
,$ is
\begin{array}{l}\require{cancel}
y^2-4y-8=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
y^2-4y-8=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-4)\pm\sqrt{(-4)^2-4(1)(-8)}}{2(1)}
\\\\=
\dfrac{4\pm\sqrt{16+32}}{22}
\\\\=
\dfrac{4\pm\sqrt{48}}{2}
\\\\=
\dfrac{4\pm\sqrt{16\cdot3}}{2}
\\\\=
\dfrac{4\pm\sqrt{(4)^2\cdot3}}{2}
\\\\=
\dfrac{4\pm4\sqrt{3}}{2}
\\\\=
\dfrac{2(2\pm2\sqrt{3})}{2}
\\\\=
\dfrac{\cancel{2}(2\pm2\sqrt{3})}{\cancel{2}}
\\\\=
2\pm2\sqrt{3}
.\end{array}
Hence, the solutions are $
\left\{ 2-2\sqrt{3},2+2\sqrt{3} \right\}
.$
