1. **State the problem:** Solve the quadratic equation $x^2 + 5x + 6 = 0$. 2. **Formula and rules:** The quadratic equation is generally solved using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the equation $ax^2 + bx + c = 0$. 3. **Identify coefficients:** Here, $a = 1$, $b = 5$, and $c = 6$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 5^2 - 4 \times 1 \times 6 = 25 - 24 = 1$$ Since $\Delta > 0$, there are two distinct real roots. 5. **Apply the quadratic formula:** $$x = \frac{-5 \pm \sqrt{1}}{2 \times 1} = \frac{-5 \pm 1}{2}$$ 6. **Find the roots:** - For $+$ sign: $$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$$ - For $-$ sign: $$x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$$ 7. **Final answer:** The solutions to the equation are $x = -2$ and $x = -3$.