To solve the quadratic equation \( 4x^2 + 2x = 10 \), we first rearrange the equation into standard form:
\[ 4x^2 + 2x - 10 = 0. \]
Next, we can utilize the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 2 \), and \( c = -10 \).
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 2^2 - 4(4)(-10) = 4 + 160 = 164. \]
Now we can plug into the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{164}}{2 \cdot 4} = \frac{-2 \pm \sqrt{164}}{8}. \]
Next, we simplify \( \sqrt{164} \):
\[ \sqrt{164} = \sqrt{4 \cdot 41} = 2\sqrt{41}. \]
Substituting this back into the equation gives:
\[ x = \frac{-2 \pm 2\sqrt{41}}{8} = \frac{-2}{8} \pm \frac{2\sqrt{41}}{8} = \frac{-1 \pm \sqrt{41}}{4}. \]
Thus, the two solutions are:
\[ x = \frac{-1 + \sqrt{41}}{4} \quad \text{and} \quad x = \frac{-1 - \sqrt{41}}{4}. \]
Now we compare these solutions with the provided options:
- \(-2 + \frac{4\sqrt{4}}{1}\) (This does not match)
- \(-1 - \frac{4\sqrt{41}}{4}\) \(\equiv \frac{-1 - \sqrt{41}}{4}\) (This matches the second solution)
- \(-2 + \frac{9\sqrt{2}}{8}\) (This does not match)
- \(-1 - \frac{9\sqrt{2}}{8}\) (This does not match)
The correct solution from the options is:
−1−41√4
This corresponds to \( \frac{-1 - \sqrt{41}}{4} \).