(x^2 + 4x)^2 + (x^2 + 4x) - 12 = 0
Пусть х^2 + 4х = у, (x^2 + 4x)^2 = y^2
y^2 + y - 12 = 0
По теореме Виета
У1 = -4, У2 = 3
1) x^2 + 4x = -4
x^2 + 4x + 4 = 0
(x + 2)^2 = 0
X1,2 = -2
2) x^2 + 4x = 3
x^2 + 4x - 3 = 0
D = b^2 - 4ac = 4^2 - 4* (-3) = 16 + 12 = 28>0
X3 = (-4 - 28^1/2) / 2 = (-4 - 2 * 7^1/2) / 2 = -2 - 7^1/2
X4 = (-4 + 28^1/2) / 2 = - 2 + 7^1/2