jmc

Completing the square in $x$ and $y,$ we get $$16(x-2{)}^2+(y+2{)}^2=64.$$Then $$\frac{(x-2{)}^2}{4}+\frac{(y+2{)}^2}{64}=1.$$Thus, $a=8$ and $b=2,$ so $c=\sqrt{a^2-b^2}=\sqrt{60}=2\sqrt{15}.$ Therefore, the distance between the foci is $2c=\boxed{4\sqrt{15}}.$