[x(x+3)][(x+1)(x+2)]=360
(x^2+3x)(x^2+3x+2)=360
设t=x^2+3x+1,则有(t-1)(t+1)=360,解得t^2=361,t1=19,t2=-19
则有x^2+3x+1=19,解得x1=3,x2=-6
或x^2+3x+1=-19,此时(x+3/2)^2+71/4=0,矛盾,舍去
所以x1=3,x2=-6