А) x² + x = 0
x (x+1) = 0
x = 0 или x+1 = 0
x = 0 или x = -1
Ответ: -1;0
б) y (y-3) + 4 (3-y) = 0
y² - 3y + 12 - 4y = 0
y² - 7y + 12 = 0
D = (-7)² - 12 * 4 = 49 - 48 = 1
y₁ = (7+1)/2 = 8/2 = 4
y₂ = (7-1)/2 = 6/2 = 3
Ответ: 3;4
в) y³ + 5y² - y - 5 = 0
y²(y + 5) - (y + 5) = 0
(y+5) (y² -1) = 0
(y+5) (y-1) (y+1) = 0
y₁ = -5
y₂ = 1
y₃ = -1
Ответ: -5 ; -1 ; 1