[Algebra]
Problem: Determine all values of $x$ for which $2+\sqrt{x-2}=x-2$.

Solution: Rearranging and then squaring both sides,

$$
\begin{aligned}
2+\sqrt{x-2} & =x-2 \\
\sqrt{x-2} & =x-4 \\
x-2 & =(x-4)^{2} \\
x-2 & =x^{2}-8 x+16 \\
0 & =x^{2}-9 x+18 \\
0 & =(x-3)(x-6)
\end{aligned}
$$

so $x=3$ or $x=6$.

We should check both solutions, because we may have introduced extraneous solutions by squaring.

If $x=3$, the left side equals $2+\sqrt{1}=3$ and the right side equals 1 , so $x=3$ must be rejected.

If $x=6$, the left side equals $2+\sqrt{4}=4$ and the right side equals 4 , so $x=6$ is the only solution.
Suppose $u=\sqrt{x-2}$.

The equation becomes $2+u=u^{2}$ or $u^{2}-u-2=0$ or $(u-2)(u+1)=0$.

Therefore, $u=2$ or $u=-1$.

But we cannot have $\sqrt{x-2}=-1$ (as square roots are always non-negative).

Therefore, $\sqrt{x-2}=2$ or $x-2=4$ or $x=6$.

Final answer: 6