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