To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we first need to identify the values of \(a\), \(b\), and \(c\):
- \(a = 3\)
- \(b = 18\)
- \(c = 24\)
Next, we will calculate the discriminant (\(b^2 - 4ac\)):
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Calculate \(b^2\): \[ b^2 = 18^2 = 324 \]
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Calculate \(4ac\): \[ 4ac = 4 \cdot 3 \cdot 24 = 288 \]
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Calculate \(b^2 - 4ac\): \[ b^2 - 4ac = 324 - 288 = 36 \]
Now we substitute \(a\), \(b\), and the discriminant into the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-18 \pm \sqrt{36}}{2 \cdot 3} \]
- Calculate \(\sqrt{36}\): \[ \sqrt{36} = 6 \]
Now substitute this back into the formula:
\[ x = \frac{-18 \pm 6}{6} \]
- This gives us two possible solutions for \(x\):
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When using the plus sign: \[ x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2 \]
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When using the minus sign: \[ x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4 \]
So, the solutions to the equation \(3x^2 + 18x + 24 = 0\) are:
Answers:
- \(x_1 = -2\)
- \(x_2 = -4\)
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