$2.13 \sqrt{x-2}=x-4$.

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$2.13 \sqrt{x-2}=x-4$.
2.13 Let's square both sides of the equation:

$$
\begin{aligned}
& (\sqrt{x-2})^{2}=(x-4)^{2} ; x-2=x^{2}-8 x+16 \\
& x^{2}-9 x+18=0 ; x_{1}=3 ; x_{2}=6
\end{aligned}
$$

Let's check the found roots by substituting them into the original equation. If $x=3$, we get $1=-1$ - an incorrect equality; if $x=6$, we get $2=2$ - a correct equality. Therefore, the given equation has a single root $x=6$.

Note that we could have first found the domain of the given equation. For this, we solve the system of inequalities

$\left\{\begin{array}{l}x-2 \geqslant 0, \\ x-4 \geqslant 0,\end{array}\right.$

from which $x \geqslant 4$. Then it is immediately clear that $x_{1}=3$ is an extraneous root of the original equation. By checking, we confirm that $x=6$ satisfies the given equation.

Answer: $x=6$.