Решите уравнение 20 корень 2x+корень x-1=3

√2х + √х-1 = 3

одз:
1)х-1≥0
х≥1
2) 2х≥0
х≥0
хэ[1;+бесконечность)

(√2х + √х-1)^2 = 3^3
2х + 2√2х(х-1) + х-1 = 9
2√2х(х-1) = 10-3х
(2√2х(х-1))^2 = (10-3х)^2
4*2х(х-1) = 100-60х+9х^2
8х^2-8х-100+60х-9х^2 = 0
-х^2+52х-100 = 0
х^2-52х+100 = 0
Д = 52^2-400 = 2704-400 = 2304 = 48^2
х1 = (52+48)/2 = 100/2 = 50
х2 = (52-48)/2 = 4/2 = 2