1. **Problem statement:** Solve the inequality $$3x^2 + 2x - 8 \leq 0$$ given that $$3x^2 + 2x - 8 = (x + 2)(3x - 4)$$. 2. **Formula and rules:** To solve quadratic inequalities, first find the roots by setting the quadratic equal to zero, then determine the intervals where the quadratic is positive or negative using a sign chart. 3. **Find roots:** Set $$3x^2 + 2x - 8 = 0$$. Given factorization: $$(x + 2)(3x - 4) = 0$$. So roots are: $$x + 2 = 0 \Rightarrow x = -2$$ $$3x - 4 = 0 \Rightarrow x = \frac{4}{3}$$ 4. **Sign analysis:** The parabola opens upwards (coefficient of $$x^2$$ is positive). - For $$x < -2$$, test point $$x = -3$$: $$(x + 2)(3x - 4) = (-3 + 2)(3(-3) - 4) = (-1)(-13) = 13 > 0$$ - For $$-2 < x < \frac{4}{3}$$, test point $$x = 0$$: $$(0 + 2)(3(0) - 4) = 2 \times (-4) = -8 < 0$$ - For $$x > \frac{4}{3}$$, test point $$x = 2$$: $$(2 + 2)(3(2) - 4) = 4 \times 2 = 8 > 0$$ 5. **Solution:** The inequality $$3x^2 + 2x - 8 \leq 0$$ holds where the expression is negative or zero, i.e., between the roots including them: $$\boxed{-2 \leq x \leq \frac{4}{3}}$$ This means all real numbers $$x$$ in this interval satisfy the inequality.