Question

помогите решить уравнения.
(х+1)(х+2)-(х-3)(х+4)=6
х3+3х²-4х-12=0

Answer 1

$(x+1)(x+2)-(x-3)(x+4)=6\x^2+3x+2-x^2-x+12=6\2x=-8\x=-4$

$x^3+3x^2-4x-12=0\x^2(x+3)-4(x+3)=0\(x^2-4)(x+3)=0\x^2-4=0 x+3=0\x^2=4 x=-3\x=^+_-2$