X*(x+1)*(x+2)*(x+3) = (x*(x+3))*((x+1)*(x+2)) = ( x^2 +3x)*( x^2 +3x+2) = = 120,
сделаем замену x^2+3x = y, тогда
y*(y+2) =120;
y^2 + 2y - 120 = 0;
D/4 = 1+120 = 121=11^2;
y1 = (-1+11) = 10;
y2 = (-1-11) = -12;
1) x^2 + 3x = 10;
x^2 + 3x - 10 = 0;
D = 9 - 4*(-10) = 49 = 7^2;
x1 = (-3+7)/2 = 4/2 = 2;
x2 = (-3-7)/2 = -10/2 = -5;
2) x^2 + 3x = -12;
x^2 + 3x + 12 = 0;
D = 9 - 4*12 < 0; в 2) решений нет.
Ответ. x1 = 2; x2 = -5;