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 :)