From Förberedande kurs i matematik 1

Square both sides of the equation so that the root sign disappears,

\displaystyle 1-x=\left( 2-x \right)^{2}\quad \Leftrightarrow \quad 1-x=4-4x+x^{2}

and then solve the resulting second-order equation by completing the square:

\displaystyle \begin{align} & x^{2}-3x+3=0 \\ & \left( x-\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+3=0 \\ & \left( x-\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{12}{4}=0 \\ & \left( x-\frac{3}{2} \right)^{2}+\frac{3}{4}=0 \\ \end{align}

As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to \displaystyle {3}/{4}\;, regardless of how \displaystyle x is chosen; so, the original root equation does not have any solutions.