jmc

Completing the square in $x$ and $y,$ we get $$(x-3{)}^2-4(y+1{)}^2=32.$$Then $$\frac{(x-3{)}^2}{32}-\frac{(y+1{)}^2}{8}=1.$$We see that $a^2=32$ and $b^2=8,$ so $c^2=a^2+b^2=40,$ and $c=2\sqrt{10}.$ Therefore, the distance between the foci is $2c=\boxed{4\sqrt{10}}.$