I have been wondering, can you prove that if we suppose in a graph (connected or not) there are no cycles of length 3 (so at least 4) and each vertex has a degree 4 or less, then the number of edges will be less than $\frac{3}{2}*n - 2$ where n is the number of vertices.
I tried using Euler's Formula, namely $v - e + f = k + 1$. If we then take into consideration the restriction on cycle length we get $2f < m$ so we can substitute and get $m < 2n - 4$, however that is not enough and also is not using the fact that each vertex has a degree 4 or less.
I'm sorry, I forgot to tell the graph has to be planar.
A non-planar counterexample to your hypothesis. the Folkman graph is quartic (4-regular, all vertices have degree 4) and has girth (minimum cycle length) of 4. This has 20 vertices and 40 edges.
Adding in the planar requirement, an infinite square grid would obviously not meet your formula, so a large square grid with the corners looped around to each other should also fail... let's try $6\times 6$, which has $36$ vertices and $64$ edges breaks your proposed limit of $\frac{3}{2}*n - 2$:
... and I could sneak in four more edges to get to $2n{-}4$ in fact, by joining the nodes adjacent to the corners also along each edge.
It's pretty clear that this construction allows a $2n{-}4$ edged triangle-free graph on $n$ nodes for a grid that is $2p\times 2q$ with $p,q>1$, nearly $4$-regular except for $8$ degree-$3$ nodes. The smallest is thus on $16$ vertices with $28$ edges:
