` (2x-1)^2 = x(x+2)`
` => 4x^2 - 4x +1 = x^2 +2x`
` => 4x^2 -4x +1 - x^2 -2x = 0`
` => 3x^2 -6x + 1 =0`
` Δ' = (-3)^2 - 3.1 = 9 - 3 = 6`
` => x_1 = (-b'+\sqrt{Δ'})/a = ( 3 + \sqrt(6))/3`
` =>x_2 = (-b'-\sqrt{Δ'})/a = ( 3 - \sqrt(6))/3`
Vậy `x ∈ { ( 3 + \sqrt(6))/3; ( 3 - \sqrt(6))/3}`