## Algebra 1

$s=4$
$\sqrt {2s+8}=s$ $(\sqrt {2s+8})^2=s^2$ $2s+8=s^2$ $2s+8-2s-8=s^2-2s-8$ $s^2-2s-8=0$ $(s+2)(s-4)=0$ $s+2=0$ $s=-2$ (we can't have the square root of a number be negative, so this answer is not valid) $s-4= 0$ $s=4$ $\sqrt {2s+8}=s$ $\sqrt {2*4+8}=4$ $\sqrt {8+8}=4$ $\sqrt {16} = 4$ $4=4$