First, we start with the equation:
\[ 4x^2 - 2x + 2 - 3x - x^2 = 0. \]
Let's combine like terms:
-
Combine the \(x^2\) terms: \(4x^2 - x^2 = 3x^2\).
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Combine the \(x\) terms: \(-2x - 3x = -5x\).
So, the equation simplifies to:
\[ 3x^2 - 5x + 2 = 0. \]
Now, we can apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \(a = 3\), \(b = -5\), and \(c = 2\).
First, calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 = (-5)^2 = 25, \] \[ 4ac = 4 \times 3 \times 2 = 24, \] \[ b^2 - 4ac = 25 - 24 = 1. \]
Now plug the values into the quadratic formula:
\[ x = \frac{-(-5) \pm \sqrt{1}}{2 \cdot 3} = \frac{5 \pm 1}{6}. \]
This gives us two potential solutions:
- \(x = \frac{5 + 1}{6} = \frac{6}{6} = 1\),
- \(x = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3}\).
Thus, the solutions are:
\[ x = \frac{2}{3} \quad \text{and} \quad x = 1. \]
In simplified form, we place the smaller value first:
\[ x = \left(\frac{2}{3}\right) \text{ and } (1). \]
Final answer:
x = (\(\frac{2}{3}\)) and (1)