1. **State the problem:** Solve the quadratic equation $$6(-4m - 5)^2 + 1(-4m - 5) - 1 = 0$$ for $m$. 2. **Substitute:** Let $x = -4m - 5$ to simplify the equation. 3. **Rewrite the equation:** It becomes $$6x^2 + x - 1 = 0$$. 4. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, solutions are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 5. **Identify coefficients:** Here, $a=6$, $b=1$, $c=-1$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 1^2 - 4 \times 6 \times (-1) = 1 + 24 = 25$$. 7. **Find $x$ values:** $$x = \frac{-1 \pm \sqrt{25}}{2 \times 6} = \frac{-1 \pm 5}{12}$$ 8. **Calculate each root:** - $$x_1 = \frac{-1 + 5}{12} = \frac{4}{12} = \frac{1}{3}$$ - $$x_2 = \frac{-1 - 5}{12} = \frac{-6}{12} = -\frac{1}{2}$$ 9. **Back-substitute for $m$:** Recall $x = -4m - 5$. For $x_1 = \frac{1}{3}$: $$-4m - 5 = \frac{1}{3}$$ $$-4m = \frac{1}{3} + 5 = \frac{1}{3} + \frac{15}{3} = \frac{16}{3}$$ $$m = \frac{\cancel{-4}m}{\cancel{-4}} = \frac{-16/3}{4} = -\frac{16}{3} \times \frac{1}{4} = -\frac{16}{12} = -\frac{4}{3}$$ For $x_2 = -\frac{1}{2}$: $$-4m - 5 = -\frac{1}{2}$$ $$-4m = -\frac{1}{2} + 5 = -\frac{1}{2} + \frac{10}{2} = \frac{9}{2}$$ $$m = \frac{\cancel{-4}m}{\cancel{-4}} = \frac{-9/2}{4} = -\frac{9}{2} \times \frac{1}{4} = -\frac{9}{8}$$ 10. **Final solutions:** $$m = -\frac{4}{3}, \quad m = -\frac{9}{8}$$