jmc

To get the standard form of the equation of the hyperbola, we complete the square in both variables: $$\begin{array}{rl}-6(x^2-4x)+5(y^2+4y)& =64\\ -6(x^2-4x+4)+5(y^2+4y+4)& =64-6(4)+5(4)\\ -6(x-2{)}^2+5(y+2{)}^2& =60\\ \frac{(y+2{)}^2}{12}-\frac{(x-2{)}^2}{10}& =1.\end{array}$$Then, the distance from each focus to the center must be $\sqrt{12+10}=\sqrt{22},$ so the distance between the two foci is $\boxed{2\sqrt{22}}.$