Question

Solve -2x^2 + 3x - 1 = -2 using the quadratic formula round the solution(s) to the nearest hundredths

Answer 1

To solve the equation \(-2x^2 + 3x - 1 = -2\), we first rearrange it into standard form, which takes the form \(ax^2 + bx + c = 0\).

Starting with the original equation:

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

Add \(2\) to both sides to get:

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

This simplifies to:

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

To put it in standard form, we can multiply the entire equation by \(-1\):

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

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

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

In this case, the coefficients are:

- \(a = 2\)
- \(b = -3\)
- \(c = -1\)

Next, we calculate the discriminant (\(b^2 - 4ac\)):

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

Now substitute the values back into the quadratic formula:

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

Next, we need to calculate the two possible values of \(x\):

1. For the positive root:

\[
x_1 = \frac{3 + \sqrt{17}}{4} \approx \frac{3 + 4.1231}{4} \approx \frac{7.1231}{4} \approx 1.7808
\]

2. For the negative root:

\[
x_2 = \frac{3 - \sqrt{17}}{4} \approx \frac{3 - 4.1231}{4} \approx \frac{-1.1231}{4} \approx -0.2808
\]

Rounding the solutions to the nearest hundredths, we get:

\[
x_1 \approx 1.78 \quad \text{and} \quad x_2 \approx -0.28
\]

Thus, the solutions to the equation are:

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