To solve the quadratic equation \( 3x^2 - 13x + 5 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation:
- \( a = 3 \)
- \( b = -13 \)
- \( c = 5 \)
Now, we calculate the values:
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Calculate the discriminant: \[ b^2 - 4ac = (-13)^2 - 4 \cdot 3 \cdot 5 = 169 - 60 = 109 \]
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Now substituting the values into the quadratic formula: \[ x = \frac{-(-13) \pm \sqrt{109}}{2 \cdot 3} \] \[ x = \frac{13 \pm \sqrt{109}}{6} \]
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Now we need to calculate \( \sqrt{109} \): \[ \sqrt{109} \approx 10.4403 \]
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Substitute back: \[ x = \frac{13 \pm 10.4403}{6} \]
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Now calculate the two possible values for \( x \):
- First value: \[ x_1 = \frac{13 + 10.4403}{6} = \frac{23.4403}{6} \approx 3.9067 \approx 3.9 \text{ (to the nearest tenth)} \]
- Second value: \[ x_2 = \frac{13 - 10.4403}{6} = \frac{2.5597}{6} \approx 0.4266 \approx 0.4 \text{ (to the nearest tenth)} \]
The two solutions rounded to the nearest tenth are:
- \( x \approx 3.9 \)
- \( x \approx 0.4 \)
Thus, the solutions are \( x = 3.9 \) and \( x = 0.4 \).