Answer
$\{-5, -1\}.$
Work Step by Step
First, exclude any x that yields 0 in a denominator
$ x\neq 3,-3$
$\displaystyle \frac{2x}{x-3}+\frac{6}{x+3}=\frac{-28}{x^{2}-9} \qquad $ ... multiply with LCD = $(x-3)(x+3)$
$ 2x(x+3)+6(x-3)=-28\qquad $ ... write as $ ax^{2}+bx+c=0$
$2x^{2}+6x+6x-18=-28$
$ 2x^{2}+12x+10=0\qquad $ ... divide with 2
$ x^{2}+6x+5=0$
We factor this trinomial by finding two factors of 5 that add to 6; they are +5 and +1.
$(x+1)(x+5)=0$
$ x=-1,\qquad x=-5\qquad $(neither was excluded)
The solution set is
$\{-5, -1\}.$
