To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula, we first identify the coefficients:
- \(a = 3\)
- \(b = 18\)
- \(c = 24\)
The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Now, we calculate the discriminant:
\[
b^2 - 4ac = 18^2 - 4 \times 3 \times 24 = 324 - 288 = 36
\]
Next, we apply the quadratic formula:
\[
x = \frac{-18 \pm \sqrt{36}}{2 \times 3}
\]
Calculating \(\sqrt{36}\):
\[
\sqrt{36} = 6
\]
Now substitute back into the formula:
\[
x = \frac{-18 \pm 6}{6}
\]
This gives us two possible solutions:
1.
\[
x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2
\]
2.
\[
x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4
\]
Therefore, the solutions are:
\[
x = -2 \quad \text{and} \quad x = -4
\]
So the final answer is:
**x = −2, x = −4**.