Пусть x^2 + 2x = t, тогда
t^2 - 2t - 3= 0
D = 4 + 12 = 16 = 4^2
t1 = (2 +4)/2 = 6/2 = 3
t2 =( 2 - 4)/2 = - 2/2 = - 1
Два случая
1) x^2 + 2x = 3
x^2 + 2x - 3 = 0
D = 4 + 12= 16 = 4^2
x1 = ( - 2 + 4)/2 = 2/2 = 1
x2 = ( - 2 - 4)/2 = - 6/2 = - 3
2) x^2 + 2x = - 1
x^2 + 2x + 1 = 0
(x + 1)^2 = 0
x + 1 = 0
x3 = - 1
Ответ
- 3; - 1; 1