To use the quadratic formula, we need to identify the values of \( a \), \( b \), and \( c \) from the quadratic equation \( 3x^2 + 18x + 24 = 0 \).
From the equation:
- \( a = 3 \)
- \( b = 18 \)
- \( c = 24 \)
Now, we will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Now, let's fill in the necessary values:
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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 the discriminant, \( b^2 - 4ac \): \[ b^2 - 4ac = 324 - 288 = 36 \]
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Now substitute these values into the quadratic formula: \[ x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3} \]
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Calculate \( \sqrt{36} \): \[ \sqrt{36} = 6 \]
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Substitute back into the quadratic formula: \[ x = \frac{-18 \pm 6}{6} \]
Now we can solve for the two potential values of \( x \):
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First solution: \[ x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2 \]
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Second solution: \[ x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4 \]
Thus, the solutions to the equation \( 3x^2 + 18x + 24 = 0 \) are \( x = -2 \) and \( x = -4 \).
If you were looking for specific answers to fill in the yellow boxes, they would be:
- \( a \) = 3
- \( b \) = 18
- \( c \) = 24
- \( b^2 = 324 \)
- \( 4ac = 288 \)
- \( b^2 - 4ac = 36 \)
- \( \sqrt{b^2 - 4ac} = 6 \)
- Solutions: \( x = -2 \) and \( x = -4 \)