Number of edges of a convex polytope with n vertices

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Homework Statement



What is the number of edges of a convex polytope with n vertices all of whose faces are triangles.


Homework Equations



# of faces + #of vertecies = # of edges + 2

The Attempt at a Solution



My reasoning is as follows:

n/3 + n = # of edges + 2

4n/3 - 2 = #of edges

This doesn't seem right though. Am I using the correct formula?
 
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Right formula. Wrong logic. Each edge meets two faces. Each face meets three edges. What's the relation between edges and faces?
 
3/2 * faces = edges?

Is the result

3/2*n - 2 = # of edges by any chance?

I just did

3/2*1/3*n + n = e + 2
 
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Yes. Three times the number of faces equals twice the number of edges. If you have to do this again pick some example like tetrahedron E=6, V=4, F=4. Octohedron E=12, V=6 and F=8. It's a good sanity check.
 
That's actually how I found it. I drew a little tetrahedron with 4 vertecies, 6 edges and 4 faces. I don't know why but i just couldn't wrap my mind around it without a drawing. Thanks again for your time.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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