Using Euler’s formula for vertices, edges and faces of a polyhedron:

V+F-E = 2

If we let the number of pentagonal face be n, then the number of faces is (n+2), the number of edges is (14+5n)/2 (since each edge belongs to two faces), and the number of vertices is (14+5n)/3 (since each vertex belongs to three faces).

(14+5n)/3 + (n+2) – (14+5n)/2 = 2

Which boils down to n = 14, so there are 14 pentagonal faces.

Below is such a polyhedron. All of the edges and all of the vertices are visible, just one heptagonal face is on the far side.