1. Stating the problem: Solve the quadratic equation $ (3x-4)^2 - 5(3x-4) + 6 = 0 $ for $x$. 2. Let $y = 3x - 4$. Then the equation becomes $y^2 - 5y + 6 = 0$. 3. Factor the quadratic in $y$: $y^2 - 5y + 6 = (y - 2)(y - 3) = 0$. 4. Set each factor equal to zero: - $y - 2 = 0 \implies y = 2$ - $y - 3 = 0 \implies y = 3$ 5. Substitute back $y = 3x - 4$: - For $y = 2$: $3x - 4 = 2 \implies 3x = 6 \implies x = 2$ - For $y = 3$: $3x - 4 = 3 \implies 3x = 7 \implies x = \frac{7}{3}$ 6. Final solutions are $x = 2$ and $x = \frac{7}{3}$.