Question

Solve -2x^2 + 3x - 1 = -2. Round to nearest hundredths

Answer 1

To solve the equation \(-2x^2 + 3x - 1 = -2\), we first move all terms to one side of the equation:

\[
-2x^2 + 3x - 1 + 2 = 0
\]

This simplifies to:

\[
-2x^2 + 3x + 1 = 0
\]

Next, we can multiply the entire equation by -1 to make the leading coefficient positive:

\[
2x^2 - 3x - 1 = 0
\]

Now we will use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our case, \(a = 2\), \(b = -3\), and \(c = -1\).

Let's calculate the discriminant (\(b^2 - 4ac\)):

\[
b^2 - 4ac = (-3)^2 - 4(2)(-1) = 9 + 8 = 17
\]

Now, substituting \(a\), \(b\), and the discriminant into the quadratic formula:

\[
x = \frac{-(-3) \pm \sqrt{17}}{2(2)} = \frac{3 \pm \sqrt{17}}{4}
\]

Now we can calculate the two possible values for \(x\):

\[
x_1 = \frac{3 + \sqrt{17}}{4}, \quad x_2 = \frac{3 - \sqrt{17}}{4}
\]

Calculating each root:

1. For \(x_1\):

\(\sqrt{17} \approx 4.123\)

\[
x_1 \approx \frac{3 + 4.123}{4} \approx \frac{7.123}{4} \approx 1.78075
\]

2. For \(x_2\):

\[
x_2 \approx \frac{3 - 4.123}{4} \approx \frac{-1.123}{4} \approx -0.28075
\]

Finally, rounding to the nearest hundredths, we get:

\[
x_1 \approx 1.78, \quad x_2 \approx -0.28
\]

Thus, the solutions are:

\[
\boxed{1.78} \quad \text{and} \quad \boxed{-0.28}
\]