Question

using the quadratic formula find the solution 3x^2 + 18x + 24 = 0

Answer 1

To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula, we first identify the coefficients \(a\), \(b\), and \(c\):

- \(a = 3\)
- \(b = 18\)
- \(c = 24\)

The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

First, we need to calculate the discriminant, \(b^2 - 4ac\):

\[
b^2 = 18^2 = 324
\]

\[
4ac = 4 \cdot 3 \cdot 24 = 288
\]

Now, substitute these into the discriminant:

\[
b^2 - 4ac = 324 - 288 = 36
\]

Since the discriminant is positive, we will have two distinct real solutions.

Now plugging in the values into the quadratic formula:

\[
x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3}
\]

Calculating \(\sqrt{36}\):

\[
\sqrt{36} = 6
\]

Thus, we have:

\[
x = \frac{-18 \pm 6}{6}
\]

Now we can calculate the two possible values for \(x\):

1. For the positive root:

\[
x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2
\]

2. For the negative root:

\[
x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4
\]

Therefore, the solutions to the equation \(3x^2 + 18x + 24 = 0\) are:

\[
x = -2 \quad \text{and} \quad x = -4
\]