1. Solve the inequality $$-6x^2 + 5x + 6 < 0$$. 2. First, find the roots of the quadratic equation $$-6x^2 + 5x + 6 = 0$$ by using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a = -6$$, $$b = 5$$, and $$c = 6$$. 3. Calculate the discriminant: $$\Delta = 5^2 - 4(-6)(6) = 25 + 144 = 169$$. 4. Find the roots: $$x = \frac{-5 \pm \sqrt{169}}{2(-6)} = \frac{-5 \pm 13}{-12}$$. 5. Calculate each root: - $$x_1 = \frac{-5 + 13}{-12} = \frac{8}{-12} = -\frac{2}{3}$$ - $$x_2 = \frac{-5 - 13}{-12} = \frac{-18}{-12} = \frac{3}{2}$$ 6. Since $$a = -6 < 0$$, the parabola opens downward, so the quadratic is less than zero between the roots: $$-\frac{2}{3} < x < \frac{3}{2}$$. 7. Solve the inequality $$\frac{2}{x} - \frac{1}{x^2} < -3$$. 8. Multiply both sides by $$x^2$$ (noting $$x \neq 0$$) to clear denominators: $$2x - 1 < -3x^2$$. 9. Rearrange to standard form: $$3x^2 + 2x - 1 < 0$$. 10. Find roots of $$3x^2 + 2x - 1 = 0$$ using quadratic formula with $$a=3$$, $$b=2$$, $$c=-1$$: $$\Delta = 2^2 - 4(3)(-1) = 4 + 12 = 16$$. 11. Roots: $$x = \frac{-2 \pm 4}{2 \cdot 3}$$ - $$x_1 = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}$$ - $$x_2 = \frac{-2 - 4}{6} = \frac{-6}{6} = -1$$ 12. Since $$a=3 > 0$$, parabola opens upward, so inequality $$3x^2 + 2x - 1 < 0$$ holds between roots: $$-1 < x < \frac{1}{3}$$. 13. Also, remember $$x \neq 0$$ because of original denominators. 14. Final solution for second inequality: $$-1 < x < 0$$ or $$0 < x < \frac{1}{3}$$. **Final answers:** - For $$-6x^2 + 5x + 6 < 0$$: $$-\frac{2}{3} < x < \frac{3}{2}$$. - For $$\frac{2}{x} - \frac{1}{x^2} < -3$$: $$-1 < x < 0$$ or $$0 < x < \frac{1}{3}$$.