Solve for all solutions of x : \( \begin{array}{l}x^{2}-4 x+4=2 \\ x=\{ \pm \sqrt{2}\} \\ x=\{-2,+2\} \\ x=\{1+\sqrt{2}, 1-\sqrt{2}\} \\ x=\{2+\sqrt{2}, 2-\sqrt{2}\}\end{array} \) \begin{tabular}{l}\( x=\left\{\begin{array}{l}x=2\end{array}\right. \) \\ \hline\end{tabular}

Let's first simplify the equation \(x^{2}-4x+4=2\). Rearranging gives us: \[ x^2 - 4x + 2 = 0. \] Now we can use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = 2\). Calculating the discriminant: \[ b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 2 = 16 - 8 = 8. \] Now substituting back into the quadratic formula: \[ x = \frac{4 \pm \sqrt{8}}{2}. \] Since \(\sqrt{8} = 2\sqrt{2}\), we get: \[ x = \frac{4 \pm 2\sqrt{2}}{2}, \] which simplifies to: \[ x = 2 \pm \sqrt{2}. \] Thus, the solutions are: \[ x = 2 + \sqrt{2}, \quad x = 2 - \sqrt{2}. \] So the final answer is: \[ x=\{2+\sqrt{2}, 2-\sqrt{2}\}. \] This would eliminate the other potential answers listed.