Solve for \( u \). \( \frac{-11 u}{u^{2}-u-30}-\frac{8}{u-6}=\frac{u}{u+5} \) If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".

To solve the equation \[ \frac{-11 u}{u^{2}-u-30}-\frac{8}{u-6}=\frac{u}{u+5}, \] we start by factoring the quadratic \(u^{2}-u-30\). This can be factored as \((u-6)(u+5)\). Now, substitute this back into the equation: \[ \frac{-11 u}{(u-6)(u+5)}-\frac{8}{u-6}=\frac{u}{u+5}. \] Next, find a common denominator for all terms. The common denominator is \((u-6)(u+5)\). Rewriting each term gives: \[ \frac{-11u}{(u-6)(u+5)} - \frac{8(u+5)}{(u-6)(u+5)} = \frac{u(u-6)}{(u-6)(u+5)}. \] Now combine the left side: \[ \frac{-11u - 8(u+5)}{(u-6)(u+5)} = \frac{u(u-6)}{(u-6)(u+5)}. \] This simplifies to: \[ \frac{-11u - 8u - 40}{(u-6)(u+5)} = \frac{u^2 - 6u}{(u-6)(u+5)}. \] Combine the left side: \[ \frac{-19u - 40}{(u-6)(u+5)} = \frac{u^2 - 6u}{(u-6)(u+5)}. \] Since both sides have the same denominator, we can set the numerators equal: \[ -19u - 40 = u^2 - 6u. \] Rearranging gives: \[ u^2 + 13u + 40 = 0. \] Next, we factor the quadratic: \[ (u + 8)(u + 5) = 0. \] This gives us the solutions: \[ u + 8 = 0 \quad \Rightarrow \quad u = -8, \] \[ u + 5 = 0 \quad \Rightarrow \quad u = -5. \] We need to check for any restrictions from denominators. The values \( u = 6 \) and \( u = -5 \) would make the denominators zero. Thus, we exclude \( u = -5 \). Therefore, the valid solution is: \[ u = -8. \]