Question

（х＋1）（х＋2）-（х-3）（х+4）=6, решите.

Answer 1

x^2+2x+x+2-x^2-x+12=6

2x+14=6

2x+14-6=0

2x+8=0

x+4=0

x=-4

Answer 2

$(x+1)(x+2)-(x-3)(x+4)=6\\x^2+2x+1x+2-(x^2-3x+4x-12)=6\\x^2+3x+2-x^2-x+12-6=0\\2x+8=0\\2x=-8\\x=-4$

Ответ: -4.