The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \(3x^2 + 18x + 24 = 0\), we can identify the coefficients as follows:
- \(a = 3\)
- \(b = 18\)
- \(c = 24\)
Now, we can use these coefficients in the quadratic formula:
- Calculate \(b^2 - 4ac\):
\[ b^2 = 18^2 = 324 \] \[ 4ac = 4 \cdot 3 \cdot 24 = 288 \] \[ b^2 - 4ac = 324 - 288 = 36 \]
- Apply the values into the quadratic formula:
\[ x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3} \]
- Calculate \(\sqrt{36}\):
\[ \sqrt{36} = 6 \]
- Substitute this back into the formula:
\[ x = \frac{-18 \pm 6}{6} \]
- Calculate the two possible values for \(x\):
First Value (lesser number):
\[ x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4 \]
Second Value (greater number):
\[ x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2 \]
Thus, the solutions to the equation \(3x^2 + 18x + 24 = 0\) are:
- \(x = -4\) (lesser number)
- \(x = -2\) (greater number)
Final answer:
- \(x = -4\) (lesser number)
- \(x = -2\) (greater number)
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