To solve the equation \( x^2 - 2x + 2 - 3x - x = 0 \), we first simplify the equation.
Combine the like terms:
\[ x^2 - 2x - 3x - x + 2 = 0 \] \[ x^2 - 6x + 2 = 0 \]
Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
In this case, \( a = 1 \), \( b = -6 \), and \( c = 2 \).
Calculate the discriminant:
\[ b^2 - 4ac = (-6)^2 - 4(1)(2) = 36 - 8 = 28 \]
Now substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[ x = \frac{-(-6) \pm \sqrt{28}}{2(1)} = \frac{6 \pm \sqrt{28}}{2} \]
We can simplify \( \sqrt{28} \):
\[ \sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7} \]
Substituting back into the formula:
\[ x = \frac{6 \pm 2\sqrt{7}}{2} \]
This can be simplified further:
\[ x = 3 \pm \sqrt{7} \]
Thus, the two solutions are:
\[ x_1 = 3 + \sqrt{7} \] \[ x_2 = 3 - \sqrt{7} \]
Since we want the smaller value first:
The answers are: \[ x = 3 - \sqrt{7} \quad \text{and} \quad x = 3 + \sqrt{7} \]
So the final boxed answer should be:
Smaller value:
\[ 3 - \sqrt{7} \]
Larger value:
\[ 3 + \sqrt{7} \]