Solve the equation given to find the value(s) of \( p \). \( \begin{array}{l}-(3 p-2)^{2}-5(3 p-2)+6= \\ -2(3 p-2)^{2}\end{array} \) INSTRUCTION: Enter your answers separated by the " ; symbol.

To solve the equation \(-(3p-2)^{2}-5(3p-2)+6=-2(3p-2)^{2}\), we first simplify both sides. Starting with the left side: \[ -(3p-2)^{2}-5(3p-2)+6 \] Let \(x = 3p - 2\). Then the equation can be rewritten as: \[ -x^2 - 5x + 6 = -2x^2 \] Next, we move all terms to one side of the equation: \[ -x^2 - 5x + 6 + 2x^2 = 0 \] This simplifies to: \[ x^2 - 5x + 6 = 0 \] Now we can factor the quadratic: \[ (x - 2)(x - 3) = 0 \] Setting each factor to zero gives: \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] Recalling that \(x = 3p - 2\), we substitute back: 1. For \(x = 2\): \[ 3p - 2 = 2 \quad \Rightarrow \quad 3p = 4 \quad \Rightarrow \quad p = \frac{4}{3} \] 2. For \(x = 3\): \[ 3p - 2 = 3 \quad \Rightarrow \quad 3p = 5 \quad \Rightarrow \quad p = \frac{5}{3} \] Thus, the values of \(p\) are: \[ \frac{4}{3} ; \frac{5}{3} \]