To solve the equation \(2x^2 + x = 5\), we first rearrange it into standard form:
\[ 2x^2 + x - 5 = 0 \]
Next, we can apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -5\).
Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4(2)(-5) = 1 + 40 = 41 \]
Now substituting into the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{41}}{2(2)} = \frac{-1 \pm \sqrt{41}}{4} \]
Calculating the two solutions:
- \(x_1 = \frac{-1 + \sqrt{41}}{4}\)
- \(x_2 = \frac{-1 - \sqrt{41}}{4}\)
Now we need the numerical values:
Calculating \(\sqrt{41} \approx 6.4\):
\[ x_1 \approx \frac{-1 + 6.4}{4} = \frac{5.4}{4} \approx 1.35 \]
\[ x_2 \approx \frac{-1 - 6.4}{4} = \frac{-7.4}{4} \approx -1.85 \]
Listing the solutions, with the smaller one first:
\[ x \approx -1.85 \quad \text{and} \quad x \approx 1.35 \]
Thus, rounding to the hundredths:
\[ x \approx -1.85 \quad \text{and} \quad x \approx 1.35 \]
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