6.300 \sqrt{x-2}+\sqrt{4-x}=x^{2}-6 x+11
Solution. $\sqrt{(x-3)+1}+\sqrt{1-(x-3)}=\left(x^{2}-6 x+9\right)+2$.

Let $y=x-3$, then $\sqrt{y+1}+\sqrt{1-y}=y^{2}+2 \Leftrightarrow$ $y^{4}+4 y^{2}+2\left(1-\sqrt{1-y^{2}}\right)=0$.

Since $1 \geq \sqrt{1-y^{2}}$, the last equation is equivalent to the system:

$$
\left\{\begin{array}{l}
y^{4}+4 y^{2}=0 \\
1=\sqrt{1-y^{2}}
\end{array} \Leftrightarrow y=0 .\right.
$$

Answer: $x=3$.