jmc

To put this equation in standard form, we complete the square in each variable: $$\begin{array}{rl}(x^2-4x)+3(y^2+10y)& =2\\ (x^2-4x+4)+3(y^2+10y+25)& =2+4+3(25)\\ (x-2{)}^2+3(y+5{)}^2& =81.\end{array}$$Dividing through by $81$ yields the standard form of the ellipse:$$\frac{(x-2{)}^2}{81}+\frac{(y+5{)}^2}{27}=1.$$It follows that the semiminor axis has length $\sqrt{27}=3\sqrt{3},$ so the minor axis has length $2\cdot 3\sqrt{3}=\boxed{6\sqrt{3}}.$