To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula, we start by identifying the coefficients:
- \(a = 3\)
- \(b = 18\)
- \(c = 24\)
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let's calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = 18^2 = 324 \] \[ 4ac = 4 \times 3 \times 24 = 288 \]
Now, calculate the discriminant:
\[ b^2 - 4ac = 324 - 288 = 36 \]
Next, we can plug in the values into the quadratic formula:
\[ x = \frac{-18 \pm \sqrt{36}}{2 \times 3} \]
The square root of 36 is 6, so we have:
\[ x = \frac{-18 \pm 6}{6} \]
Now, we can evaluate the two possible values for \(x\):
-
\(x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2\)
-
\(x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4\)
Thus, the solutions are:
\[ x = -2 \quad \text{and} \quad x = -4 \]
From the provided options, the correct response would be:
x = -2, x = -4