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:
-
\(x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2\)
-
\(x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4\)
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).