9. Let $x, y \in \mathbf{R}$ satisfy
$$
x-6 \sqrt{y}-4 \sqrt{x-y}+12=0 \text {. }
$$

Then the range of values for $x$ is $\qquad$
9. $[14-2 \sqrt{13}, 14+2 \sqrt{13}]$.

From the problem, we have
$$
\begin{array}{l}
(\sqrt{x-y}-2)^{2}+(\sqrt{y}-3)^{2}=1 . \\
\text { Let } \sqrt{x-y}-2=\cos \theta, \sqrt{y}-3=\sin \theta \text {. Then } \\
x=(2+\cos \theta)^{2}+(3+\sin \theta)^{2} \\
=14+\sqrt{52} \sin (\theta+\varphi)\left(\sin \varphi=\frac{2}{\sqrt{13}}\right) .
\end{array}
$$

Therefore, $14-2 \sqrt{13} \leqslant x \leqslant 14+2 \sqrt{13}$.