решите уравнение:

x^2-3x^2-4x+12=0

 

помогите плиииз)

 

x^3 - 3x^2 - 4x + 12 = 0
x^3 - 4x - (3x^2 - 12) = 0
x(x^2 - 4) - 3(x^2 - 4) = 0
(x^2 - 4)(x - 3) = 0
(x - 2)*(x + 2)*(x - 3) = 0
x1 = 2,
x2 = -2,
x3 = 3