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To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula, we first need to identify the coefficients \(a\), \(b\), and \(c\):
- \(a = 3\)
- \(b = 18\)
- \(c = 24\)
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Now let's calculate the discriminant \(b^2 - 4ac\):
\[
b^2 = 18^2 = 324
\]
\[
4ac = 4 \cdot 3 \cdot 24 = 288
\]
Now, calculate the discriminant:
\[
b^2 - 4ac = 324 - 288 = 36
\]
Now, plug the values into the quadratic formula:
\[
x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3}
\]
Simplifying further:
\[
x = \frac{-18 \pm 6}{6}
\]
This gives us two possible solutions:
1. \(x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2\)
2. \(x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4\)
Thus, the solutions are \(x = -2\) and \(x = -4\).
The correct response is:
**x = -2, x = -4** (or \(x\) equals negative 2, \(x\) equals negative 4).