Решите уравнение (x^2-x+1)^2-10(x-4)(x+3)-109=0

(x^2 - x + 1)^2 - 10(x - 4)(x + 3) - 109 = 0;

(x^2 - x + 1)^2 - 10(x^2 + 3x - 4x - 12) - 109 = 0;

(x^2 - x + 1)^2 - 10(x^2 - x - 12) - 109 = 0;

(x^2 - x + 1)^2 - 10(x^2 - x + 1 - 13) - 109 = 0;

введем новую переменную x^2 - x + 1 = y;

y^2 - 10(y - 13) - 109 = 0;

y^2 - 10y + 130 - 109 = 0;

y^2 - 10y + 21 = 0;

D = b^2 - 4ac;

D = (-10)^2 - 4 * 1 * 21 = 100 - 84 = 16; √D = 4;

x = (-b ± √D)/(2a);

y1 = (10 + 4)/2 = 7; 

y2 = (10 - 4)/2 = 3.

Выполним обратную подстановку:

1) x^2 - x + 1 = 7;

x^2 - x + 1 - 7 = 0;

x^2 - x - 6 = 0;

D = (-1)^2 - 4 * 1 * (-6) = 1 + 24 = 25; √D = 5;

x1 = (1 + 5)/2 = 3;

x2 = (1 - 5)/2 = -2.

2) x^2 - x + 1 = 3;

x^2 - x + 1 - 3 = 0;

x^2 - x - 2 = 0;

D = (-1)^2 - 4 * 1 * (-2) = 1 + 8 = 9; √D = 3;

x3 = (1 + 3)/2 = 2;

x4 = (1 - 3)/2 = -1.

Ответ. -2; -1; 2; 3.