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:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
First, calculate the discriminant:
\[ b^2 - 4ac = 18^2 - 4 \cdot 3 \cdot 24 \] \[ = 324 - 288 \] \[ = 36 \]
Next, since the discriminant is positive, we will have two real solutions. Now plug the values into the quadratic formula:
\[ x = \frac{{-18 \pm \sqrt{36}}}{{2 \cdot 3}} \] \[ = \frac{{-18 \pm 6}}{{6}} \]
Calculate the two solutions:
-
\( x = \frac{{-18 + 6}}{6} = \frac{{-12}}{6} = -2 \)
-
\( x = \frac{{-18 - 6}}{6} = \frac{{-24}}{6} = -4 \)
So the solutions are:
\[ x = -2 \quad \text{and} \quad x = -4 \]
Thus, the response is:
x=−2, x=−4 (or x equals negative 2 , x equals negative 4)