## Elementary Algebra

{$-2,2$}
Rearranging the equation and taking the square root on both sides, we obtain: $2x^{2}-8=0$ $2x^{2}=8$ $x^{2}=\frac{8}{2}$ $x^{2}=4$ $\sqrt {x^{2}}=\pm\sqrt 4$ $x=\pm2$ Therefore, the solution set is {$-2,2$}.