To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula, we first identify the coefficients \(a\), \(b\), and \(c\) from the standard form \(ax^2 + bx + c = 0\).
Here:
- \(a = 3\)
- \(b = 18\)
- \(c = 24\)
The quadratic formula is given by:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
Now, we can calculate the values step by step:
- Calculate the discriminant \(D = b^2 - 4ac\):
\[ D = 18^2 - 4 \times 3 \times 24 = 324 - 288 = 36 \]
-
Since the discriminant \(D = 36\) is positive, there will be two distinct real solutions.
-
Now substitute \(a\), \(b\), and \(D\) into the quadratic formula:
\[ x = \frac{{-18 \pm \sqrt{36}}}{2 \times 3} \]
- Calculate \(\sqrt{36}\):
\[ \sqrt{36} = 6 \]
- Substitute this back into the equation:
\[ x = \frac{{-18 \pm 6}}{6} \]
- Now calculate the two possible values for \(x\):
- For the "+" case:
\[ x = \frac{{-18 + 6}}{6} = \frac{{-12}}{6} = -2 \]
- For the "−" case:
\[ x = \frac{{-18 - 6}}{6} = \frac{{-24}}{6} = -4 \]
Thus, the solutions to the equation \(3x^2 + 18x + 24 = 0\) are:
\[ x = -2, \quad x = -4 \]
So the correct response is:
x = -2, x = -4