Since $\angle C = \dfrac{\pi}{2}$, $\sin B = \sin\left(\dfrac{\pi}{2} - A\right) = \cos A \implies \sin^2 A + \sin^2 B = 1$. By Vieta's Formulas, we have $\sin A + \sin B = \dfrac{7n - 20}{5n + 8}$ and $\sin A \sin B = \dfrac{120}{5n + 8}$. Now, $1 = \sin^2 A + \sin^2 B = (\sin A + \sin B)^2 - 2\sin A \sin B = \left(\dfrac{7n - 20}{5n + 8}\right)^2 - 2\left(\dfrac{120}{5n + 8}\right)$ $= \dfrac{(7n - 20)^2 - 240(5n + 8)}{(5n + 8)^2} \implies (7n - 20)^2 - 240(5n + 8) = (5n + 8)^2$ $\implies 49n^2 -1480n - 1520 = 25n^2 + 80n + 64 = 0 \implies 24n^2 - 1560n - 1584 = 0$ $\implies n^2 - 65n - 66 = 0 \implies (n - 66)(n + 1) = 0 \implies n = -1, 66$. However, since $A, B \in \left(0, \dfrac{\pi}{2}\right)$, $\sin A + \sin B = \dfrac{7n - 20}{5n + 8} > 0$. This holds true only for $n = 66$, which is our answer.