You probably can't recall how to solve this as it doesn't have a classical form.
First identify the degree of the equation. The LHS includes a square root, i.e. degree $\frac12$; the RHS is of the first degree. This is not a polynomial equation, you need to reshape it for easier handling.
Get rid of the square root by isolating it and squaring:
Doing that, you possibly introduce an alien solution, to be rejected later by remembering that $x-2\ge0$.
Next, get rid of the absolute value. You do it by splitting in two cases:
$$x<0\implies -x=(x-2)^2\\x\ge0\implies x=(x-2)^2.$$
Rearranging, you now have two nice quadratic equations for which standard formulas are available:
$$x<0\implies x^2-3x+4=0\\x\ge0\implies x^2-5x+4=0.$$
After solving, check the constrains.
You could as well get rid of the absolute value by squaring, as $|x|^2=x^2$, but this would lead you to a scary quartic equation.