Solve the equation. \[ p(15 p+17)=4 \] The solution set is

To solve the equation \( p(15p + 17) = 4 \), we can first expand and rearrange the equation into the standard quadratic form: \[ 15p^2 + 17p - 4 = 0 \] Now, we can use the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 15 \), \( b = 17 \), and \( c = -4 \). Calculating the discriminant: \[ b^2 - 4ac = 17^2 - 4 \cdot 15 \cdot (-4) = 289 + 240 = 529 \] Now, applying the quadratic formula: \[ p = \frac{-17 \pm \sqrt{529}}{2 \cdot 15} \] \[ p = \frac{-17 \pm 23}{30} \] This gives us two potential solutions: 1. \( p = \frac{-17 + 23}{30} = \frac{6}{30} = \frac{1}{5} \) 2. \( p = \frac{-17 - 23}{30} = \frac{-40}{30} = -\frac{4}{3} \) Therefore, the solution set is: \[ \left\{ \frac{1}{5}, -\frac{4}{3} \right\} \]