#### Answer

$$x=-9$$

#### Work Step by Step

$$\frac{4}{(x-2)(x+3)}-\frac{1}{(x+2)(x-2)}=\frac{2}{(x+3)(x+2)}$$
$$((x-2)(x+3)(x+2))(\frac{4}{(x-2)(x+3)}-\frac{1}{(x+2)(x-2)})=((x-2)(x+3)(x+2))(\frac{2}{(x+3)(x+2)})$$
$$4(x+2)-(x+3)=2(x-2)$$
$$4x+8-x-3=2x-4$$
$$3x+5=2x-4$$
$$x=-9$$
-9 is not an extraneous solution; $x=-9$.