1. **State the problem:** Solve the inequality $5x^2 - 13x - 6 > 0$ using case analysis. 2. **Formula and approach:** To solve a quadratic inequality $ax^2 + bx + c > 0$, first find the roots of the quadratic equation $ax^2 + bx + c = 0$. These roots divide the number line into intervals. We test each interval to determine where the inequality holds. 3. **Find the roots:** Solve $5x^2 - 13x - 6 = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=-13$, $c=-6$. Calculate the discriminant: $$\Delta = (-13)^2 - 4 \times 5 \times (-6) = 169 + 120 = 289$$ Calculate the roots: $$x = \frac{-(-13) \pm \sqrt{289}}{2 \times 5} = \frac{13 \pm 17}{10}$$ So, $$x_1 = \frac{13 - 17}{10} = \frac{-4}{10} = -0.4$$ $$x_2 = \frac{13 + 17}{10} = \frac{30}{10} = 3$$ 4. **Intervals:** The roots divide the real line into three intervals: - $(-\infty, -0.4)$ - $(-0.4, 3)$ - $(3, \infty)$ 5. **Test each interval:** Choose test points in each interval to check the sign of $5x^2 - 13x - 6$. - For $x = -1$ (in $(-\infty, -0.4)$): $$5(-1)^2 - 13(-1) - 6 = 5 + 13 - 6 = 12 > 0$$ - For $x = 0$ (in $(-0.4, 3)$): $$5(0)^2 - 13(0) - 6 = -6 < 0$$ - For $x = 4$ (in $(3, \infty)$): $$5(4)^2 - 13(4) - 6 = 80 - 52 - 6 = 22 > 0$$ 6. **Conclusion:** The inequality $5x^2 - 13x - 6 > 0$ holds for $$x \in (-\infty, -0.4) \cup (3, \infty)$$ **Final answer:** $$\boxed{x < -0.4 \text{ or } x > 3}$$