To solve the equation \(x^2 + 6x + 9 = 0\) using the quadratic formula, we first identify the coefficients \(a\), \(b\), and \(c\) from the standard form of a quadratic equation \(ax^2 + bx + c = 0\):
- \(a = 1\)
- \(b = 6\)
- \(c = 9\)
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Now, we will calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0 \]
Since the discriminant is 0, there will be exactly one real solution (a repeated root). Now substituting the values into the quadratic formula:
\[ x = \frac{-6 \pm \sqrt{0}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{-6}{2} = -3 \]
Thus, the solution to the equation \(x^2 + 6x + 9 = 0\) is:
\[ \boxed{-3} \]