设（x-3）=Y 就是说先用换元
(x-3)²-（x-3）=12(x-3)^2-(x-3)-12=0[(x-3)-4][(x-3)+3]=0(x-7)(x)=0x1=0x2=7
(x-3)²-（x-3）=12(x-3)²-（x-3）-12=0(x-3-4)(x-3+3)=0x(x-7)=0x1=0 x2=7
