Answer
Solution set: $\displaystyle \{-3,-\frac{4}{3}\}$
Work Step by Step
$g(a)=20$
$\displaystyle \frac{5}{a+2}+\frac{25}{a^{2}+4a+4}=20$
... recognize a perfect square: $a^{2}+4a+4=a^{2}+2\cdot 2\cdot a+2^{2}$
$\displaystyle \frac{5}{a+2}+\frac{25}{(a+2)^{2}}=20$
...Multiply the equation with the LCD=$(a+2)^{2}$
$ 5(a+2)+25=20(a+2)^{2}\qquad$ ... simplify (distribute)
$5a+10+25=20(a^{2}+4a+4)$
$ 5a+35=20a^{2}+80a+80\qquad$ ... divide all with 5
$ a+7=4a^{2}+16a+16\qquad$ ... add $-a-7$
$0=4a^{2}+15a+9$
We want to factor the RHS. Search for factors of ac whose sum is b...
... we find $12$ and $3$, factors of $4\times 9=36$ whose sum is $15$
$ 0=4a^{2}+12a+3a+9\qquad$ ... factor in pairs
$0=4a(a+3)+3(a+3)$
$0=(a+3)(4a+3)$
By the zero product principle,
$a\displaystyle \in\{-3,-\frac{4}{3}\}, $ ... both values satisfy (*).
Solution set: $\displaystyle \{-3,-\frac{4}{3}\}$
