1. **State the problem:** Solve the quadratic equation $4x^2 + 11x + 6 = 0$. 2. **Formula used:** The quadratic formula is given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. 3. **Identify coefficients:** Here, $a = 4$, $b = 11$, and $c = 6$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 11^2 - 4 \times 4 \times 6 = 121 - 96 = 25$$ 5. **Apply the quadratic formula:** $$x = \frac{-11 \pm \sqrt{25}}{2 \times 4} = \frac{-11 \pm 5}{8}$$ 6. **Find the two solutions:** - For the plus sign: $$x = \frac{-11 + 5}{8} = \frac{-6}{8} = \frac{\cancel{\! -6}}{\cancel{8}} = -\frac{3}{4}$$ - For the minus sign: $$x = \frac{-11 - 5}{8} = \frac{-16}{8} = \frac{\cancel{\! -16}}{\cancel{8}} = -2$$ 7. **Final answer:** The solutions to the equation are $$x = -\frac{3}{4} \quad \text{and} \quad x = -2$$