jmc

First, we divide both sides by 7, to get $$\frac{x^2}{140}+\frac{y^2}{28}=1.$$Thus, $a^2=140$ and $b^2=28,$ so $c^2=a^2-b^2=140-28=112.$ Thus, $c=\sqrt{112}=4\sqrt{7},$ so the distance between the foci is $2c=\boxed{8\sqrt{7}}.$