Whenever an expression of the type $|f(x)|$ appears, you should separate two cases: $f(x)\geq0$ and $f(x)\leq0$.
In your case, because you have an expression $|x-2|$, you should separate the cases when $x-2\geq0$ (i.e. when $x\geq2$) and when $x-2\leq0$ (i.e. when $x\leq2$).
$x\geq2$. In this case, $x-2\geq0$ meaning the equation $|x-2|=2-x$ becomes $x-2=2-x$ which simplifies to $2x=4$ and $x=2$. So, the first equation in case $1$ holds if and only if $x=2$.
The second equation is even simpler. Because $x\geq 2$, it is clear that $x+2\geq 4>0$, so $|x+2|=x+2$ and the second equation becomes $x+2=x+2$ which is always true.
So, in Case 1, $x=2$ is the only solution that satisfies both equations.
$x\leq 2$ means $|x-2|=2-x$ is always true because $|x-2|=-(x-2)=2-x$. So, the first equation is always true.
The second equation is unclear, and you have to split two additional cases:
$x\geq -2$. Follow the steps from Case 1
$x\leq -2$. Follow the steps from Case 1.