possible polyhedra from euler's formula I'm not very clear with the euler's formula, and I couldn't find it anywhere. I'm sorry if it is a double post.
F + V - E = 2
Is the euler's formula.
If the equation balances, is it polyhedra all the time??
What if it was something like
5 vertices 11 faces and 14 edges??
What shape would it be??
 A: Just because a triple of numbers $(V, E, F)$ satisfies $V + F - E = 2$, it won't automatically be the face vector of a polyhedron. For example, there's no polyhedron in $\Bbb R^3$ with face vector $(3, 4, 3)$, as there are no polyhedra with only $3$ vertices, at least $4$ are necessary. There are other inequalities constraining valid polytopes as well, although I can't think of any at the moment (things telling you that $V$ vertices can only be used to make so many faces, for example). These formulas are all necessary conditions a polyhedron must satisfy, generally they aren't sufficient to guarantee a polyhedron exists.
In the case of $(V, E, F) = (5, 14, 11)$ in particular, such a polyhedron can't exist. To maximize $F$, our polyhedron must have all triangular faces. In this case, each face must contain exactly $3$ edges, and each edge is contained in exactly $2$ faces, so that $F = \frac{2}{3}E$. Consequently every polytopes with $F$ faces and $E$ edges is subject to $F \leq \frac{2}{3} E$, but your numbers don't satisfy that inequality.

Bonus thoughts on Euler's Formula:
Euler's formula just tells you about a single combinatorial aspect of a polyhedron. It doesn't give you a full picture, combinatorially, and it certainly doesn't tell you all there is to know about the polyhedron 'geometrically'.
For example, start with a cube, whose f-vector is $(8, 12, 6)$ as it has $8$ vertices, $12$ edges, and $6$ faces. Now chop off a corner to obtain a new polyhedron with f-vector $(10, 15, 7)$. This polyhedron has an extra triangular face where the missing vertex was, $3$ additional edges, and $2$ new vertices. (This operation is known as truncation, specifically vertex truncation, as we can chop off edges, too).
If we want to chop off a corner of this new polyhedron, several things can happen. 
We could chop off a vertex of the original that was connected to the missing vertex. We could chop off a vertex of the original cube that was directly opposite the missing vertex. We could chop off a vertex of the original cube that was neither adjacent to, nor opposite, our missing vertex. Finally, we could chop off one of our new vertices.
In all cases, we obtain a polytope whose f-vector is $(12, 18, 8)$, but could be any one of $4$ combinatorial equivalence classes of polyhedra. Euler's formula alone can't distinguish between them, however.
Going one step further, even if two polytopes are combinatorially equivalent (in the case of cutting off opposite corners of the cube, say), our resulting polytopes may not be geometrically equivalent, in this case depending on how 'deep' each of the cuts are. 
So, Euler's formula is a powerful tool, absolutely! But there are many more degrees of freedom in a polyhedron, than just knowing how many vertices, edges, and faces there are. 
A: Intro
There are some other constraints on $E,V,F$, regardless of whether the polyhedron is convex or not. In fact these hold as long as the polyhedron is simple (has no holes).
If the polyhedron is simple -- this includes all convex polyhedra, then Euler's formula applies in its original form ($F + V - E = 2$); otherwise the right-hand side becomes $2 - 2g$ where $g$ is the genus of the polyhedron (e.g. see https://plus.maths.org/content/eulers-polyhedron-formula).

General Polyhedra
Label the $i$-th vertex $V_i$ ($1 \le i \le m$), which is the junction of $k_i \ge 3$ edges for every $i$.
Label the $j$-th face $F_j$ ($1 \le j \le n$), which is a polygon with $p_j \ge 3$ vertices for every $j$.
Let $E_i$ be the contribution of $V_i$ to the total number of edges, so $E_i = k_i / 2$ (since each edge is counted twice) and
$E = \sum\limits_{i=1}^{m}{E_i} = \sum\limits_{i=1}^{m}{\frac{k_i}{2}} \ge \frac{3m}{2}$ (since $k_i \ge 3$). Then

$E \ge \frac{3}{2} V$

Let $E_j$ be the contribution of $F_j$ to the total number of edges, so $E_j = p_j / 2$ (since each edge joins two faces) and
$E = \sum\limits_{j=1}^{n}{E_j} = \sum\limits_{i=1}^{n}{\frac{p_j}{2}} \ge \frac{3n}{2}$ (since $p_j \ge 3$). Then

$E \ge \frac{3}{2} F$


Simple Polyhedra Only
Since $F + V = E + 2$ and $F \le \frac{2}{3}E$ and $V \le \frac{2}{3}E$, we also need
$F \ge \frac{1}{3}E + 2 \iff E \le 3(F-2)$ and
$F \ge \frac{1}{3}V + 2 \iff E \le 3(V-2)$
Combining with previous results:

$ \frac{3}{2}V \le E \le 3(V-2)$

and

$ \frac{3}{2}F \le E \le 3(F-2)$

A: If would not be regular, but a type of polyhedron all the same.
