To solve the equation \(-2x^2 + 3x - 1 = -2\), we first rearrange it into standard quadratic form \(ax^2 + bx + c = 0\).
Starting with: \[ -2x^2 + 3x - 1 + 2 = 0 \] \[ -2x^2 + 3x + 1 = 0 \]
Now, we can use the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = -2\), \(b = 3\), and \(c = 1\).
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Calculate \(b^2 - 4ac\): \[ b^2 - 4ac = 3^2 - 4(-2)(1) = 9 + 8 = 17 \]
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Substitute \(a\), \(b\), and \(b^2 - 4ac\) into the quadratic formula: \[ x = \frac{-3 \pm \sqrt{17}}{2(-2)} = \frac{-3 \pm \sqrt{17}}{-4} \]
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Now calculate the two possible solutions: \[ x_1 = \frac{-3 + \sqrt{17}}{-4}, \quad x_2 = \frac{-3 - \sqrt{17}}{-4} \]
Calculating each one:
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For \(x_1\): \[ \sqrt{17} \approx 4.123 \] \[ x_1 = \frac{-3 + 4.123}{-4} = \frac{1.123}{-4} \approx -0.28075 \]
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For \(x_2\): \[ x_2 = \frac{-3 - 4.123}{-4} = \frac{-7.123}{-4} \approx 1.78075 \]
Now round the solutions:
- \(x_1 \approx -0.28\)
- \(x_2 \approx 1.78\)
Thus, the smaller solution first: \[ x \approx -0.28 \quad \text{or} \quad 1.78 \]
Final rounded solutions: \[ x = -0.28 \quad \text{or} \quad 1.78 \]