√2х-1+(2х-1)=2 решите уравнение

[latex]\sqrt{2x-1}+2x-1=2 \\ (\sqrt{2x-1})^2=(2-2x+1)^2 \\ 2x-1=(3-2x)^2 \\ 2x-1=9-12x+4x^2 \\ 2x-1-9+12-4x^2=0 \\ 14x-10-4x^2=0\\ 2x^2-7x+5=0 \\ x_{1,2} = \frac{-(-7)+-\sqrt{(-7)^2-4*2*5}}{2*2} = \frac{7+-3}{4} \\ x_1 = 1 \\ x_2= \frac{5}{2} [/latex]