#### Answer

$y=2\pm2\sqrt{3}$

#### Work Step by Step

$y^{2}-8=4y$
Take the $4y$ to the left side of the equation:
$y^{2}-4y-8=0$
Use the quadratic formula to solve this equation. The formula is $y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this equation, $a=1$, $b=-4$ and $c=-8$
Substitute:
$y=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-8)}}{2(1)}=\dfrac{4\pm\sqrt{16+32}}{2}=...$
$...=\dfrac{4\pm\sqrt{48}}{2}=\dfrac{4\pm4\sqrt{3}}{2}=2\pm2\sqrt{3}$