In $\triangle ABC$, if $AB=x+2$, $BC=x$, $AC=x-2$, and $C = 120^\circ$, then $x=5$ I need help with this Law of cosines problem:

Prove that when $AB = x + 2,\ BC = x,\ AC = x - 2\;$ and $\;C = 120^\circ$ while $ABC$ is a triangle, then $x = 5$.

I need help to get to that answer. With law of cosines I got this
$$(x + 2)^2 = x^2 + (x-2)^2 -2x^2(x-2)\cos120^\circ$$
Then I solved the equation and I got that the solutions are $0,\ 1 + \sqrt{33}/2,\ 1 - \sqrt{33}/2$
I don't understand what I'm doing wrong.

It was a silly mistake. Already fixed it.
$(x + 2)^2 = x^2 + (x-2)^2 -2x(x-2)\cos 120°$
