Answer
$x=-4$
Work Step by Step
$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^{2}-4}$
Factor the denominator of the second fraction of the right side of the equation:
$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x-2)(x+2)}$
Multiply the whole equation by $(x-2)(x+2)$:
$(x+2)(x+5)=5(x-2)+28$
$x^{2}+7x+10=5x-10+28$
$x^{2}+7x+10=5x+18$
Take all terms to the left side:
$x^{2}+7x+10-5x-18=0$
Simplify:
$x^{2}+2x-8=0$
Solve by factoring:
$(x+4)(x-2)=0$
Our two solutions are:
$x=-4$ and $x=2$
If we check our answers, we'll see that the equation is undefined when we substitute $x=2$, so, our only solution is $x=-4$
