I assume tg(x) is tan(x) = sin(x)/cos(x).
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This is a formula due to Euler: v = vertices, f = faces, e = edges.
Imagine a cube, for example. It has 8 vertices, 6 faces and 12 edges, so v - e + f is 8 - 12 + 6 = 2
A tetrahedron: v = 4, f = 4, e = 6: v - e + f: 4 - 6 + 4 = 2
When x = 2/3 then (1/4)x2 - 5x + 7 becomes (1/4)(2/3)2 - 5*(2/3) + 7 → 1/9 - 10/3 + 7 → (1 - 30 + 63)/9 → 34/9 → 3 7/9
First notice this can be written as (2x)3 + 3(2x)2(3y) + 3(2x)(3y)2 + (3y)3
This is more easily recognisable as (2x + 3y)3
There is also the trivial solution x= 0
We have
$$e^{i\pi}=-1$$
This is one form of Euler's well known relationship.
$$e^{i\pi}\rightarrow \cos{\pi}+i\times\sin{\pi}\rightarrow -1+i\times0\rightarrow-1$$
It's probably simpler here just to cancel the √85 terms in the numerator and denominator, so that you are immediately left with -7/6.
Here's yet another approach!
What you say is correct Sir_Emo. Never-the-less, this question (together with the corresponding ones of sin-1 and cos-1) occurs frequently, and usually refers to the arctan (arcsin, arccos). Of course, there isn't enough information in the question as it stands to determine unequivocally which is meant here!
Very good fiora, though all the final denominators should be 325 (you have inadvertently written 235 for two of them).
+33667 
