1. The problem is to find the values of $x$ for the quadratic equation $x^2 + 5x + 6 = 0$. 2. The formula used to solve quadratic equations is 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. For the equation $x^2 + 5x + 6 = 0$, the coefficients are $a=1$, $b=5$, and $c=6$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = 5^2 - 4 \times 1 \times 6 = 25 - 24 = 1$$ 5. Since $\Delta > 0$, there are two real solutions. 6. Substitute values into the quadratic formula: $$x = \frac{-5 \pm \sqrt{1}}{2 \times 1} = \frac{-5 \pm 1}{2}$$ 7. Calculate the two solutions: - For $+$ sign: $$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$$ - For $-$ sign: $$x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$$ 8. Therefore, the values of $x$ are $-2$ and $-3$.