Question

2x2+x=5

Answer to the hundredths. List smaller solution first.
x=

and

Answer 1

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:

1. \(x_1 = \frac{-1 + \sqrt{41}}{4}\)
2. \(x_2 = \frac{-1 - \sqrt{41}}{4}\)

Now we need the numerical values:

Calculating \(\sqrt{41} \approx 6.4\):

1.
\[
x_1 \approx \frac{-1 + 6.4}{4} = \frac{5.4}{4} \approx 1.35
\]

2.
\[
x_2 \approx \frac{-1 - 6.4}{4} = \frac{-7.4}{4} \approx -1.85
\]

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:

\[
x \approx -1.85 \quad \text{and} \quad x \approx 1.35
\]