[tex] - {x}^{2} + x + 6 > 0 \\ {x}^{2} - x - 6 < 0 \\ \\ {x}^{2} - x - 6 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ x_{1} + x_{2} =1 \\ x_{1} x_{2} = - 6 \\ x_{1} =3 \\ x_{2} = - 2 \\ {x}^{2} - x - 6 = (x - 3)(x + 2) \\ \\ (x - 3)(x + 2) < 0 \\ + + + ( - 2) - - - (3) + + + \\ otvet \: \: \: x \:\epsilon \: ( - 2; \: 3)[/tex]

Целые решения: - 1 ; 0 ; 1 ; 2