1. **State the problem:** Solve the quadratic equation $x^2 + 3x + 2 = 0$. 2. **Formula and rules:** The quadratic formula to solve $ax^2 + bx + c = 0$ is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=3$, and $c=2$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 1 \times 2 = 9 - 8 = 1$$ Since $\Delta > 0$, there are two distinct real roots. 4. **Find the roots:** $$x = \frac{-3 \pm \sqrt{1}}{2 \times 1} = \frac{-3 \pm 1}{2}$$ 5. **Evaluate each root:** - For $+$ sign: $$x = \frac{-3 + 1}{2} = \frac{-2}{2} = -1$$ - For $-$ sign: $$x = \frac{-3 - 1}{2} = \frac{-4}{2} = -2$$ 6. **Final answer:** The solutions to the equation are $x = -1$ and $x = -2$.