To solve the quadratic equation \(x^2 + 6x + 9 = 0\) using the quadratic formula, we can recall that the quadratic formula is given by:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
In this equation, \(a\), \(b\), and \(c\) are the coefficients from the general form of a quadratic equation \(ax^2 + bx + c = 0\). For the given equation \(x^2 + 6x + 9 = 0\), the coefficients are:
- \(a = 1\)
- \(b = 6\)
- \(c = 9\)
Now, let's calculate the discriminant \(b^2 - 4ac\):
\[ b^2 - 4ac = 6^2 - 4(1)(9) = 36 - 36 = 0 \]
Since the discriminant is zero, there will be exactly one real solution. Now, substituting the values of \(a\), \(b\), and the discriminant into the quadratic formula:
\[ x = \frac{{-6 \pm \sqrt{0}}}{2(1)} \]
This simplifies to:
\[ x = \frac{{-6}}{2} = -3 \]
Thus, the solution to the equation \(x^2 + 6x + 9 = 0\) is:
\[ \boxed{-3} \]
This indicates that \(-3\) is a double root of the equation.
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