Solve for \( w \) \[ \frac{7}{(w+1)(w-7)}=5+\frac{6}{w-7} \]

To solve the equation \[ \frac{7}{(w+1)(w-7)} = 5 + \frac{6}{w-7}, \] we first need a common denominator on the right-hand side. The common denominator here would be \((w - 7)\): \[ 5 + \frac{6}{w - 7} = \frac{5(w - 7) + 6}{w - 7} = \frac{5w - 35 + 6}{w - 7} = \frac{5w - 29}{w - 7}. \] Now we rewrite the original equation: \[ \frac{7}{(w + 1)(w - 7)} = \frac{5w - 29}{w - 7}. \] Next, we will cross-multiply to eliminate the fractions: \[ 7(w - 7) = (5w - 29)(w + 1). \] Expanding both sides yields: \[ 7w - 49 = (5w - 29)(w + 1). \] Now, let’s expand the right-hand side: \[ 7w - 49 = 5w^2 + 5w - 29w - 29 = 5w^2 - 24w - 29. \] Rearranging the equation gives: \[ 0 = 5w^2 - 24w - 29 - 7w + 49, \] which simplifies to: \[ 0 = 5w^2 - 31w + 20. \] Now, we can apply the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5, b = -31, c = 20 \): Calculating the discriminant: \[ b^2 - 4ac = (-31)^2 - 4 \cdot 5 \cdot 20 = 961 - 400 = 561. \] Next, the roots are calculated: \[ w = \frac{31 \pm \sqrt{561}}{10}. \] Thus, the solutions for \( w \) are: \[ w = \frac{31 + \sqrt{561}}{10} \quad \text{and} \quad w = \frac{31 - \sqrt{561}}{10}. \]