6. The minimum value of $\sqrt{x^{2}-6 x+10}+\sqrt{x^{2}+6 x+13}$ is ( ).
(A) 6
(B) $6+\frac{\sqrt{3}}{2}$
(C) $3 \sqrt{5}$
(D) $2 \sqrt{10}$
6.( C).
$x$ is a real number. Establish a Cartesian coordinate system, and take $A(x, 0), B(3, -1), C(-3, 2)$.
By the distance formula between two points, we get
$$
A B=\sqrt{(x-3)^{2}+1^{2}}=\sqrt{x^{2}-6 x+10} \text {. }
$$

And $A C=\sqrt{(x+3)^{2}+2^{2}}=\sqrt{x^{2}+6 x+13}$.
When $x=1$, we have $A B+A C=\sqrt{5}+2 \sqrt{5}=3 \sqrt{5}$.
$$
\begin{aligned}
\text { The original expression } & =A B+A C \geqslant B C \\
& =\sqrt{[3-(-3)]^{2}+[(-1)-2]^{2}}=3 \sqrt{5} .
\end{aligned}
$$

Therefore, the minimum value of the original expression is $3 \sqrt{5}$.