Решить уравнение (x-3) (x^2+2x+4) = x^2+5x-24

(x - 3) (x^2 + 2x + 4) = x^2 + 5x - 24

x^2 + 5x - 24 = 0

По теореме Виета х_1 = - 8, х_2 = 3

Тогда x^2 + 5x - 24 = (x + 8) (x - 3)

(x - 3) (x^2 + 2x + 4) - (x + 8) (x - 3) = 0

(x - 3) (x^2 + 2x + 4 - x - 8) = 0

1) x - 3 = 0 x_1 = 3

2) x^2 + x - 4 = 0

D = b^2 - 4ac = 1^2 - 4*1 * (-4) = 1 + 16 = 17

x_2 = (-b + VD) / 2a = (-1 + V17) / 2

x_3 = 9-b - VD) / 2a = (-1 - V17) / 2