Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $60\%$ of the area of the hexagon. What is the sum of all possible values of $r$?

1. $[ACE]=\frac{\sqrt3}4(1+r+r^2)$ and $[CDE]=\frac{\sqrt3}4r$, and $[ABCDEF]=[ACE]+3[CDE]=\frac{\sqrt3}4(1+4r+r^2)$.
2. Setting $[ACE]=0.6\,[ABCDEF]$: $1+r+r^2=0.6(1+4r+r^2)$, i.e. $0.4r^2-1.4r+0.4=0$, or $r^2-3.5r+1=0$.
3. By Vieta's, the sum of roots is $\boxed{\dfrac72}$.