Denote: $F$ - the number of the faces; $V$ - the number of the vertices; $E$ - the number of the edges. We use the Euler's polyhedron formula: $F+V=E+2$. $E=\dfrac{4V}{2}=2V\Longrightarrow V=\dfrac{E}{2}$. $F+V=E+2\Longrightarrow F=\dfrac{E}{2}+2\quad(1)$. Denote $f_1, f_2,\dots, f_F$ the faces and let be $e_i$ the number of edges of the face $f_i, 1\le i\le F$. Each edge belongs to exactly $2$ faces. Results: $e_1+e_2+\dots+e_F=2E\quad(2)$. For the triangular faces $f_k$ we have $e_k=3$ and for the other faces $f_j$ we have $e_j\ge 4$. Using this result in the relation $2$, we obtain: $2E=e_1+e_2+\dots+e_F\ge 3m+4(F-m)$. Replacing in the relation $(1)$ results: $F\ge\dfrac{3m}{4}+F-m+2\Longrightarrow m\ge8$.