Why do I get one extra wrong solution when solving $2-x=-\sqrt{x}$? I'm trying to solve this equation: 
$$2-x=-\sqrt{x}$$
Multiply by $(-1)$:
$$\sqrt{x}=x-2$$
power of $2$:
$$x=\left(x-2\right)^2$$
then:
$$x^2-5x+4=0$$
and that means:
$$x=1, x=4$$



But $x=1$ is not a correct solution to the original equation.

Why have I got it? I've never got a wrong solution to an equation before. What is so special here?
This is because the equation $\;\sqrt x=x-2$  is not equivalent to $x=(x-2)^2$, but to
$$x=(x-2)^2\quad\textbf{and}\quad x\ge 2.$$
Remember $\sqrt x$, when it is defined, denotes the non-negative square root of $x$, hence in the present case, $x-2 \ge 0$, i.e. $x$ must be at least $2$.
Overshoot, with subsequent back-tracking to the destination, is sometimes the easiest path, a common example being the fact that, due to tricky lane changes and / or confusing roadway signage, it could well be easier to overshoot the destination and then wend your way back to it.  The phenomenon of such overshoot is just a special case of the fact that a larger footprint is easier than a smaller footprint. (cf: ‘collateral damage’, and cf: “I would have written a shorter letter, but I didn’t have the time.” -- Blaise Pascal) An example of mathematically necessary overshoot is the fact that the vertical coordinate of the brachistochrone overshoots the vertical coordinate of the destination, and, even more famously, that ‘the shortest distance between two truths in the real domain often passes through the complex domain’ – the passing through the complex domain being a stratospheric instance of overshoot, two examples of which are the explanation of the radius of convergence of 1/(1 + x^2), and the proof of the fact that the product of two numbers, each representable as the sum of two squares, is itself representable as the sum of two squares. A very common example is the squaring of both sides of an equation (in order to remove a radical), which may introduce an ‘extraneous’ root. You then ‘pull back’ to the actual solution set by inspecting each root of the second equation individually.