Answer
$a=\{-5,-1\}$
Work Step by Step
In factored form, the given equation, $
\dfrac{28}{9-a^2}=\dfrac{2a}{a-3}+\dfrac{6}{a+3}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{28}{-(a^2-9)}=\dfrac{2a}{a-3}+\dfrac{6}{a+3}
\\\\
\dfrac{28}{-(a+3)(a-3)}=\dfrac{2a}{a-3}+\dfrac{6}{a+3}
\\\\
-\dfrac{28}{(a+3)(a-3)}=\dfrac{2a}{a-3}+\dfrac{6}{a+3}
.\end{array}
Multiplying both sides by the $LCD=
(a+3)(a-3)
,$ then
\begin{array}{l}\require{cancel}
1(-28)=(a+3)(2a)+(a-3)(6)
\\\\
-28=2a^2+6a+6a-18
\\\\
-2a^2-6a-6a-28+18=0
\\\\
-2a^2-12a-10=0
\\\\
\dfrac{-2a^2-12a-10}{-2}=\dfrac{0}{-2}
\\\\
a^2+6a+5=0
\\\\
(a+5)(a+1)=0
\\\\
a=\{-5,-1\}
.\end{array}
Upon checking, both solutions satisfy the original equation. Hence, $
a=\{-5,-1\}
.$
