Exercise: Show that any deltahedron has an even number of faces.
In a deltahedron with E edges and F faces, E = (3/2) F. So F must be even, as E is an integer.
To derive the equation, imagine a pile of loose triangles, before they are joined together. As each triangle has three sides, there are three times as many triangle-edges as triangles. When we join the triangles together, two triangle-edges become one polyhedron edge, giving E = (3/2) F.
You can show in a similar manner that in any polyhedron, the total number of odd-edged faces is even. In other words, if you add the number of triangles, plus the number of pentagons, plus the number of heptagons, ... , you get an even number.