There are actually three roots for this equation, since you have a cubic (degree 3) polynomial.
You have 128 - 2x^3 = 0, or x^3 - 64 = 0. This can be factored as (x-4)(x^2 + 4x + 16) = 0.
So you have x-4 = 0 and x^2 + 4x + 16 = 0. For x-4=0, it is fairly obvious that x = 4 is a solution. For x^2 + 4x + 16, you may use the quadratic formula to find the two other roots/solutions.