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

To solve the equation \( 10 p^{2}+4 p+77=9 \), first, let's set it to zero by subtracting 9 from both sides: \[ 10 p^{2} + 4 p + 77 - 9 = 0 \] \[ 10 p^{2} + 4 p + 68 = 0 \] Next, we can apply the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 10 \), \( b = 4 \), and \( c = 68 \). Calculating the discriminant: \[ b^2 - 4ac = 4^2 - 4 \cdot 10 \cdot 68 = 16 - 272 = -256 \] Since the discriminant is negative, this means the quadratic equation has no real solutions and instead has complex solutions. To find the complex solutions, we continue with the formula: \[ p = \frac{-4 \pm \sqrt{-256}}{20} \] \[ p = \frac{-4 \pm 16i}{20} \] \[ p = \frac{-1}{5} \pm \frac{4i}{5} \] Thus, the solutions are \( p = -\frac{1}{5} + \frac{4i}{5} \) and \( p = -\frac{1}{5} - \frac{4i}{5} \).