To solve the quadratic equation \( x^2 + 6x + 9 = 0 \) using the quadratic formula, we start by identifying the coefficients from the standard form of a quadratic equation \( ax^2 + bx + c = 0 \).
Here,
- \( a = 1 \)
- \( b = 6 \)
- \( c = 9 \)
The quadratic formula is given by:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
- 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 one real root (a double root).
-
Now substitute \( b \) and the result from the discriminant into the quadratic formula:
\[ x = \frac{{-6 \pm \sqrt{0}}}{2 \cdot 1} = \frac{{-6 \pm 0}}{2} \]
- Simplifying gives:
\[ x = \frac{{-6}}{2} = -3 \]
Thus, the solution to the equation \( x^2 + 6x + 9 = 0 \) is:
\[ \boxed{-3} \]
This indicates that the quadratic has a double root at \( x = -3 \).