olve this equation and select the choice that ontains all of the solutions. \[ x^{2}-2 x+2=0 \]

To solve the equation \( x^{2}-2x+2=0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -2 \), and \( c = 2 \). Plugging in these values gives: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = \frac{2 \pm 2i}{2} = 1 \pm i \] Thus, the solutions are \( x = 1 + i \) and \( x = 1 - i \). The choice that contains all the solutions is: \( \{1 + i, 1 - i\} \). Remember, when you’re dealing with complex numbers, they always come in pairs of conjugates when derived from real coefficients!