To solve the quadratic equation \(-2x^2 + 3x - 1 = -2\), we first rearrange it into the standard form \(ax^2 + bx + c = 0\).
Starting with the original equation:
\[ -2x^2 + 3x - 1 = -2 \]
we can add 2 to both sides to get:
\[ -2x^2 + 3x - 1 + 2 = 0 \]
which simplifies to:
\[ -2x^2 + 3x + 1 = 0 \]
Next, we multiply the entire equation by -1 to simplify it further:
\[ 2x^2 - 3x - 1 = 0 \]
Now, we identify \(a = 2\), \(b = -3\), and \(c = -1\). We will use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
First, we calculate the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot (-1) = 9 + 8 = 17 \]
Now, we substitute \(a\), \(b\), and the discriminant into the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{17}}{2(2)} = \frac{3 \pm \sqrt{17}}{4} \]
Now we compute the two potential solutions:
- For the positive solution:
\[ x_1 = \frac{3 + \sqrt{17}}{4} \]
- For the negative solution:
\[ x_2 = \frac{3 - \sqrt{17}}{4} \]
Now we approximate \(\sqrt{17} \approx 4.123\). Using this approximation:
- Calculate \(x_1\):
\[ x_1 \approx \frac{3 + 4.123}{4} = \frac{7.123}{4} \approx 1.78075 \text{ (rounds to } 1.78 \text{)} \]
- Calculate \(x_2\):
\[ x_2 \approx \frac{3 - 4.123}{4} = \frac{-1.123}{4} \approx -0.28075 \text{ (rounds to } -0.28 \text{)} \]
Thus, the solutions rounded to the nearest hundredths are:
Smaller solution: \(-0.28\)
Larger solution: \(1.78\)
Final answer:
\[ \boxed{-0.28} \text{ and } \boxed{1.78} \]