1. **State the problem:** Solve the inequality $$(3x - 1)(2x - 3) \geq -2$$ and express the solution in interval notation. 2. **Rewrite the inequality:** Add 2 to both sides to get everything on one side: $$ (3x - 1)(2x - 3) + 2 \geq 0 $$ 3. **Expand the product:** $$ (3x)(2x) - (3x)(3) - (1)(2x) + (1)(3) + 2 \geq 0 $$ $$ 6x^2 - 9x - 2x + 3 + 2 \geq 0 $$ $$ 6x^2 - 11x + 5 \geq 0 $$ 4. **Solve the quadratic inequality:** First, find the roots of the quadratic equation: $$ 6x^2 - 11x + 5 = 0 $$ Use the quadratic formula: $$ x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 6 \cdot 5}}{2 \cdot 6} = \frac{11 \pm \sqrt{121 - 120}}{12} = \frac{11 \pm 1}{12} $$ 5. **Calculate the roots:** $$ x_1 = \frac{11 - 1}{12} = \frac{10}{12} = \frac{5}{6} $$ $$ x_2 = \frac{11 + 1}{12} = \frac{12}{12} = 1 $$ 6. **Determine the intervals:** The parabola opens upwards (coefficient of $x^2$ is positive), so the quadratic is $\geq 0$ outside the roots and $< 0$ between the roots. 7. **Write the solution:** $$ (-\infty, \frac{5}{6}] \cup [1, \infty) $$ **Final answer:** The solution to the inequality is $$(-\infty, \frac{5}{6}] \cup [1, \infty)$$.