令x²+x=t则t(t-14)+24=0t²-14t+24=0(t-2)(t-12)=0t=2或t=12所以x²+x=2或12若x²+x=2则（x+2）(x-1)=0x=-2或x=1若x²+x=12则(x+4)(x-3)=x=-4或x=3答：x=-2或x=1或x=-4或x=3 总共4个答案...
（X²+X）（X²+X—14）+24=0
令x²+x=t则t(t-14)+24=0t²-14t+24=0(t-2)(t-12)=0t=2或t=12所以x²+x=2或12若x²+x=2则（x+2）(x-1)=0x=-2或x=1若x²+x=12则(x+4)(x-3)=x=-4或x=3答：x=-2或x=1或x=-4或x=3 总共4个答案...