jmc

Factoring the quadratic in the numerator does not look pleasant, so we go ahead and multiply through by the denominator to get $$\begin{array}{rl}{r}^2-3r-17& =(r+4)(2r+7)\\ r^2-3r-17& =2r^2+15r+28\\ r^2+18r+45& =0\\ (r+3)(r+15)& =0\end{array}$$Therefore the solutions are $r=-3$ and $r=-15$ which have a difference of $\boxed{12}$.