1. **State the problem:** Solve the quadratic equation $x^2 + 9x + 14 = 0$. 2. **Recall the quadratic formula:** For an equation $ax^2 + bx + c = 0$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=9$, and $c=14$ in this case. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 9^2 - 4 \times 1 \times 14 = 81 - 56 = 25$$ Since $\Delta > 0$, there are two real solutions. 4. **Apply the quadratic formula:** $$x = \frac{-9 \pm \sqrt{25}}{2 \times 1} = \frac{-9 \pm 5}{2}$$ 5. **Find the two solutions:** - For the plus sign: $$x = \frac{-9 + 5}{2} = \frac{-4}{2} = -2$$ - For the minus sign: $$x = \frac{-9 - 5}{2} = \frac{-14}{2} = -7$$ 6. **Final answer:** The solutions to the equation are $x = -2$ and $x = -7$.