Use a common denominator: $\frac{15 - 2(x+2)}{x^2-4} = 1$. After moving the $1$ to the left side, notice that you can factor the numerator:
\[\frac{(x-5)(x+3)}{x^2-4} = 0\]
The roots of this equation are $\boxed{\textbf{(D)} \quad -3, 5}.$
The root(s) of $\frac {15}{x^2 - 4} - \frac {2}{x - 2} = 1$ is (are): 
$\textbf{(A)}\ -5\text{ and }3\qquad \textbf{(B)}\ \pm 2\qquad \textbf{(C)}\ 2\text{ only}\qquad \textbf{(D)}\ -3\text{ and }5\qquad \textbf{(E)}\ 3\text{ only}$