1. **State the problem:** Solve the quadratic equation $$x^2 + 9x + 18 = 0$$ using the quadratic formula. 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 $a$, $b$, and $c$ are coefficients from the equation. 3. **Identify coefficients:** Here, $a = 1$, $b = 9$, and $c = 18$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 9^2 - 4 \times 1 \times 18 = 81 - 72 = 9$$ 5. **Apply the quadratic formula:** $$x = \frac{-9 \pm \sqrt{9}}{2 \times 1} = \frac{-9 \pm 3}{2}$$ 6. **Find the two solutions:** - For the plus sign: $$x = \frac{-9 + 3}{2} = \frac{-6}{2} = -3$$ - For the minus sign: $$x = \frac{-9 - 3}{2} = \frac{-12}{2} = -6$$ 7. **Final solution set:** $$\boxed{\{-6, -3\}}$$ Note: The solution set given in the problem statement {6, 3} is incorrect; the correct roots are negative.