2-4у+у^2=-1. помогите пожалуйста
2-4y+y^2 = -1 y^2 - 4y + 2 + 1 = 0 y^2 - 4y + 3 = 0 D = (-4)^2 -4*3 = 16-12 = 4 √4 = 2 x1 = (4-2)/2 = 1 x2 = (4+2)/2 = 3 Ответ: х1=1 ; х2 = 3