1. **State the problem:** Solve the quadratic equation $h^2 + 5h + 4 = 0$ for $h$. 2. **Recall the quadratic formula:** For any quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $$h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients. 3. **Identify coefficients:** Here, $a = 1$, $b = 5$, and $c = 4$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 5^2 - 4 \times 1 \times 4 = 25 - 16 = 9$$ Since $\Delta > 0$, there are two distinct real roots. 5. **Find the roots:** $$h = \frac{-5 \pm \sqrt{9}}{2 \times 1} = \frac{-5 \pm 3}{2}$$ 6. **Calculate each root:** - For the plus sign: $$h = \frac{-5 + 3}{2} = \frac{-2}{2} = -1$$ - For the minus sign: $$h = \frac{-5 - 3}{2} = \frac{-8}{2} = -4$$ 7. **Final answer:** The solutions to the equation are $$h = -1 \quad \text{and} \quad h = -4$$