1. **State the problem:** Solve the quadratic equation $x^2 + 8x + 15 = 0$. 2. **Recall the quadratic formula:** For any quadratic equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=8$, and $c=15$ in this problem. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 8^2 - 4 \times 1 \times 15 = 64 - 60 = 4$$ Since $\Delta > 0$, there are two distinct real roots. 4. **Apply the quadratic formula:** $$x = \frac{-8 \pm \sqrt{4}}{2 \times 1} = \frac{-8 \pm 2}{2}$$ 5. **Find the two solutions:** - For the plus sign: $$x = \frac{-8 + 2}{2} = \frac{-6}{2} = -3$$ - For the minus sign: $$x = \frac{-8 - 2}{2} = \frac{-10}{2} = -5$$ 6. **Final answer:** The solutions to the equation are $x = -3$ and $x = -5$.