1. **State the problem:** Solve the quadratic equation $4x^2 + 4x - 5 = 0$ using the quadratic formula. 2. **Recall the quadratic formula:** For an equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients. 3. **Identify coefficients:** Here, $a = 4$, $b = 4$, and $c = -5$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 4^2 - 4 \times 4 \times (-5) = 16 + 80 = 96$$ 5. **Apply the quadratic formula:** $$x = \frac{-4 \pm \sqrt{96}}{2 \times 4} = \frac{-4 \pm \sqrt{96}}{8}$$ 6. **Simplify the square root:** $$\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$$ 7. **Substitute back:** $$x = \frac{-4 \pm 4\sqrt{6}}{8}$$ 8. **Simplify the fraction by canceling common factor 4:** $$x = \frac{\cancel{4}(-1 \pm \sqrt{6})}{\cancel{4} \times 2} = \frac{-1 \pm \sqrt{6}}{2}$$ 9. **Final solutions:** $$x_1 = \frac{-1 + \sqrt{6}}{2}, \quad x_2 = \frac{-1 - \sqrt{6}}{2}$$ These are the two solutions to the quadratic equation.