Answer
There is no solution for $\frac{a^2+2a}{a^2-4}\leq2$
Work Step by Step
$\frac{a^2+2a}{a^2-4}\leq2$
Distribute the 2 into the brackets
$2(a^2-4)\geq a^2+2a$
$a^2-2a-8\geq0$
$(z+2)(z-4)\geq0$
$p\leq -2$or $p \geq 4$
but z can not be in $(-\infty,-2) \cup (2,\infty)$
Thus, there is no solution for $\frac{a^2+2a}{a^2-4}\leq2$
