quadratic equation precalculus from Stewart, Precalculus, 5th, p56, Q. 79
Find all real solutions of the equation
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$
my solution
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$
$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{-2\pm6}{2}$$
$$x=-4\text{ or }2$$
official answer at the back of the book has only one real solution of $-4$
where did I go wrong?

$$\frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$$
multiply both sides with $(x-2)(x+2)\neq 0$ or $x\neq 2,x\neq -2$
we get
$$(x+5)(x+2)=5(x-2)+28,x\neq 2,-2$$
$$x^2+2x-8=0,x\neq 2,-2$$
$$x^2-2x+4x-8=0,x\neq 2,-2$$
$$x(x-2)+4(x-2)=0,x\neq 2,-2$$
$$(x-2)(x+4)=0,x\neq 2,-2$$
so $x+4=0$ or $x=-4\neq 2,-2$ is unique solution because from $x-2=0$ follow solution $x=-2$ that is prohibited
