jmc

To put this equation in standard form, we complete the square in each variable: $$\begin{array}{rl}(x^2+6x)+2(y^2+4y)& =15\\ (x^2+6x+9)+2(y^2+4y+4)& =15+9+2(4)=32\\ (x+3{)}^2+2(y+2{)}^2& =32.\end{array}$$Dividing through by $32$ puts this equation in standard form: $$\begin{array}{rl}& \\ \frac{(x+3{)}^2}{32}+\frac{(y+2{)}^2}{16}& =1.\end{array}$$It follows that the semimajor axis has length $\sqrt{32}=4\sqrt{2},$ so the major axis has length $2\cdot 4\sqrt{2}=\boxed{8\sqrt{2}}.$