To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula, we will 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, let's calculate the discriminant:
\[ b^2 - 4ac = 18^2 - 4 \cdot 3 \cdot 24 = 324 - 288 = 36 \]
Now that we have the discriminant, we can plug it into the formula:
\[ x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3} \]
Calculating the square root of 36:
\[ \sqrt{36} = 6 \]
Now substituting that back into the equation:
\[ x = \frac{-18 \pm 6}{6} \]
Now, we will calculate the two possible values for \(x\):
- Using the plus sign:
\[ x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2 \]
- Using the minus sign:
\[ x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4 \]
Thus, the solutions to the equation \(3x^2 + 18x + 24 = 0\) are:
\[ x = -2 \quad \text{and} \quad x = -4 \]
So the correct response is:
x=−2, x=−4
(x equals negative 2 , x equals negative 4)