Answer
{$\frac{1}{2}$}
Work Step by Step
$2(a^{2}+2)-8=-2a(a-2)-5$
$2a^{2}+4-8=-2a^{2}+4a-5$
$2a^{2}+2a^{2}-4a+4-8+5=0$
$4a^{2}-4a+1=0$
$4a^{2}-2a-2a+1=0$
$2a(2a-1)-1(2a-1)=0$
$(2a-1)(2a-1)=0$
$(2a-1)=0$ and $(2a-1)=0$
$a=\frac{1}{2}$ and $a=\frac{1}{2}$
Therefore, the solution set is {$\frac{1}{2}$}.
