We have that \(K_{3,4}\) has 7 vertices and 12 edges (each vertex in the group of 3 has degree 4). Then by Euler's formula we have that \(7 - 12 + f = 2\) so if the graph were planar, it would have \(f = 7\) faces. However, since the girth of the graph is 4 (there are no cycles of length 3) we get that \(4f \le 2e\text{.}\) But this would mean that \(28 \le 24\text{,}\) a contradiction.