Answer
$\{-5,-1\}$.
Work Step by Step
Replace $x$ with $a$ in the given functions.
$f(a)=\frac{6}{a+3},g(a)=\frac{2a}{a-3}$ and $h(a)=-\frac{28}{a^2-9}$
The given expression is
$\Rightarrow (f+g)(a)=h(a)$
Or we can write using the sum of functions formula.
$\Rightarrow f(a)+g(a)=h(a)$
Substitute all fucntions.
$\Rightarrow \frac{6}{a+3}+\frac{2a}{a-3}=-\frac{28}{a^2-9}$.....(1)
In order to clear fractions we determine the Least Common Denominator (LCD) of all denominators.
Factor the denominator on the right hand side.
Factor $a^2-9$
$\Rightarrow a^2-3^2$
Use the special formula $A^2-B^2=(A+B)(A-B)$.
$\Rightarrow (a+3)(a-3)$
Back substitute into the equation.
$\Rightarrow \frac{6}{a+3}+\frac{2a}{a-3}=-\frac{28}{(a+3)(a-3)}$
Multiply both sides by the LCD $(a+3)(a-3)$
$\Rightarrow (a+3)(a-3)\left (\frac{6}{a+3}+\frac{2a}{a-3}\right )=(a+3)(a-3)\left(-\frac{28}{(a+3)(a-3)}\right )$
Use the distributive property.
$\Rightarrow (a+3)(a-3)\left (\frac{6}{a+3}\right )+(a+3)(a-3)\left (\frac{2a}{a-3}\right )=(a+3)(a-3)\left(-\frac{28}{(a+3)(a-3)}\right )$
Cancel common factors.
$\Rightarrow 6(a-3)+2a(a+3)=-28$
Use the distributive property.
$\Rightarrow 6a-18+2a^2+6a=-28$
Add like terms.
$\Rightarrow 2a^2+12a-18=-28$
Add $28$ to both sides.
$\Rightarrow 2a^2+12a-18+28=-28+28$
Simplify.
$\Rightarrow 2a^2+12a+10=0$
Factor out $2$:
$\Rightarrow 2(a^2+6a+5)=0$
Divide both sides by $2$:
$\Rightarrow a^2+6a+5=0$
Rewrite the middle term $6a$ as $5a+a$.
$\Rightarrow a^2+5a+a+5=0$
Group terms.
$\Rightarrow (a^2+5a)+(a+5)=0$
Factor each group.
$\Rightarrow a(a+5)+(a+5)=0$
Factor out $(a+5)$.
$\Rightarrow (a+5)(a+1)=0$
Set each factor equal zero.
$\Rightarrow a+5=0$ or $a+1=0$
Isolate $a$.
$\Rightarrow a=-5$ or $a=-1$
The solution set is $\{-5,-1\}$.
Note: We have to verify if the solutions we got check the equation (1). The equation is defined for all real values of $a$ except the zeros of the denominators, which are $-3$ and $3$. Since the solutions we got are $-5$ and $-1$, therefore different than $-3$ and $3$, they are correct,
