Geometry

We can use the law of cosines to solve this problem. 

First,let's say

\(AB = a = x\\ AC = b + 2x+1 \\ BC = c = 3x+5 \\ ∠BAC = 120°\)

The law of cosine states that 

From the problem, we can write \(cos(120) = \frac{(3x+5)^2+(2x+1)^2 - x^2}{2(2x+1)(3x+5)}\)

Expanding out everything and simplifying, we get \(18x^{2}+47x+31=0\)

Using the quadratic formula, we have \(x=\frac{-{47}\pm \sqrt{{47}^{2}-4\cdot {18}\cdot {31}}}{2\cdot{18}}\)

We get \(x=-\frac{47}{36}\pm \sqrt{23}\cdot (\frac{1}{36}i)\)

So \(x=-\frac{47}{36}+\sqrt{23}\cdot (\frac{1}{36}i)\)

Thanks! :)

Edit: THIS IS MY 200th answer! Not flexing...but noice :)