Prove that simple graph $G$ is orientable I am a student in an introductory class for graph theory and this question is on our first homework assignment and I don't know where to start to prove the following:


  Suppose that $n \geq 3$ and a simple graph G has $n$ vertices and at
  least $(n^2 - 3n + 6)/2$ edges. Prove that G is orientable.


Thank you for your time. Any hints or help will be greatly appreciated.

EDIT: A graph g is orientable if it is the underlying graph of a strongly connected digraph. A digraph is strongly connected if for all vertices $v$ and $w$ there exists a walk $v-w$.
By Robin's Theorem, a graph is orientable if and only if it is connected and has no bridge. 

So try to prove:

Step 1: Any graphs with $(n^2 - 3n + 6)/2$ edges is connected.

Step 2: Any graphs with $(n^2 - 3n + 6)/2$ edges cannot have bridges.

Step 1 is pretty standard and easy. For step 2, assume by contradiction that $G$ has a bridge $b$ then, by erasing this bridge you get a graph with two disjoint components, one with $k$ vertices and the other one with $n-k$ vertices.

What is the maximum number of vertices in each component?
First thing which come to my mind is to suppose that this is non-orientable graph.

So if we suppose that this is non-orientable, then it follows to 2 cases (maybe more):

1) It is a path graph.
2) It has a bridges in it (so cycles, as well).

What is the number of edges it should have, if it is a path graph?
What if it has bridges+cycles?

Keep in mind it is simple graph, so no loops and multiple edges.