jmc

We could cross-multiply, but that doesn't look like much fun. Instead, we first factor the quadratic, which gives us $$\frac{(t-8)(t+7)}{t-8}=\frac{3}{t+5}.$$Canceling the common factor on the left gives $$t+7=\frac{3}{t+5}.$$Multiplying both sides by $t+5$ gives us $(t+7)(t+5)=3$. Expanding the product on the left gives $t^2+12t+35=3$, and rearranging this equation gives $t^2+12t+32=0$. Factoring gives $(t+4)(t+8)=0$, which has solutions $t=-4$ and $t=-8$. The greatest of these solutions is $\boxed{-4}$.