Answer
The solutions are $x=-1$ and $x=-5$
Work Step by Step
$\dfrac{2x}{x-3}+\dfrac{6}{x+3}=-\dfrac{28}{x^{2}-9}$
Factor the denominator of the fraction on the right side of the equation:
$\dfrac{2x}{x-3}+\dfrac{6}{x+3}=-\dfrac{28}{(x-3)(x+3)}$
Multiply the whole equation by $(x-3)(x+3)$:
$(x-3)(x+3)\Big[\dfrac{2x}{x-3}+\dfrac{6}{x+3}=-\dfrac{28}{(x-3)(x+3)}\Big]$
$2x(x+3)+6(x-3)=-28$
$2x^{2}+6x+6x-18=-28$
Take all terms to the left side:
$2x^{2}+6x+6x-18+28=0$
Simplify:
$2x^{2}+12x+10=0$
Take out common factor $2$ from the left side:
$2(x^{2}+6x+5)=0$
Take the $2$ to divide the right side:
$x^{2}+6x+5=\dfrac{0}{2}$
$x^{2}+6x+5=0$
Solve by factoring:
$(x+5)(x+1)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x+5=0$
$x=-5$
$x+1=0$
$x=-1$
The solutions are $x=-1$ and $x=-5$
