23) \( 10 p^{2}+4 p+77=9 \)

To solve the equation \( 10 p^{2}+4 p+77=9 \), first, we rearrange it: 1. Subtract 9 from both sides to set it to zero: \[ 10 p^{2} + 4 p + 77 - 9 = 0 \] which simplifies to: \[ 10 p^{2} + 4 p + 68 = 0 \] 2. Next, apply the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 10 \), \( b = 4 \), and \( c = 68 \): - Calculate the discriminant: \[ b^2 - 4ac = 4^2 - 4 \cdot 10 \cdot 68 = 16 - 272 = -256 \] Since the discriminant is negative, there are no real solutions. Instead, we can find complex solutions. 3. Plugging into the quadratic formula gives: \[ p = \frac{-4 \pm \sqrt{-256}}{2 \cdot 10} = \frac{-4 \pm 16i}{20} = \frac{-1 \pm 4i}{5} \] Thus, the solutions are: \[ p = \frac{-1 + 4i}{5} \quad \text{and} \quad p = \frac{-1 - 4i}{5} \]