To solve the quadratic equation \( 4x^2 + 2x = 10 \), we can first rearrange it into standard form:
\[ 4x^2 + 2x - 10 = 0 \]
Next, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, \( a = 4 \), \( b = 2 \), and \( c = -10 \). Now, we can substitute these values into the formula:
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Calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 = 2^2 = 4 \] \[ 4ac = 4 \cdot 4 \cdot (-10) = -160 \] \[ b^2 - 4ac = 4 - (-160) = 4 + 160 = 164 \]
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Now, calculate the square root of the discriminant: \[ \sqrt{164} \approx 12.81 \]
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Substitute back into the quadratic formula: \[ x = \frac{-2 \pm 12.81}{2 \cdot 4} \] \[ x = \frac{-2 \pm 12.81}{8} \]
Now we can calculate the two possible solutions:
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For \( x = \frac{-2 + 12.81}{8} \): \[ x = \frac{10.81}{8} \approx 1.35 \]
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For \( x = \frac{-2 - 12.81}{8} \): \[ x = \frac{-14.81}{8} \approx -1.85 \]
Thus, rounding to the nearest hundredth, the solutions are approximately \( x \approx 1.35 \) and \( x \approx -1.85 \).
Among the choices you provided:
- \( x = -0.77 \)
- \( x = 1.15 \)
- \( x = 1.04 \)
- \( x = -1.85 \)
The correct solution to the quadratic equation rounded to the nearest hundredth is \( x = -1.85 \).