2. Let real numbers $x, y$ satisfy
$$
x^{2}-8 x+y^{2}-6 y+24=0 \text {. }
$$

Then the maximum value of $x-2 y$ is $\qquad$ $\therefore$
2. $\sqrt{5}-2$.

From $x^{2}-8 x+y^{2}-6 y+24=0$, we get
$$
(x-4)^{2}+(y-3)^{2}=1 \text {. }
$$

When the line $x-2 y=k$ is tangent to the circle, it is the desired maximum value.
$$
\text { By } \frac{|4-6-k|}{\sqrt{1+2^{2}}}=1
$$
$\Rightarrow k=-2 \pm \sqrt{5}$ (negative value discarded).