这个配方对不对啊，人才请进来

2x²-4x+6=0
2[x²-2x+3]=0
2[(x-1)²-1+3]=0
2[(x-1)²+2]=0
2(x-1)²+4=0

2x²-4x+6=0 两边同时除2
x²-2x+3=0
(x-1)²+1+3=0
(x-1)²+4=0

第三步错了！应该是这样：2[(x-1)²-1+3]=0；2[(x-1)²+2]=0

2x2-4x+6=0

2[x2-2x+3]=0

2[(x-1)2+1-1+3]=0 <-错了 因为(x-1)2 ＝ x2 -2x +1 所以2〔(x-1)2－1＋3〕＝ 0

2[(x-1)2+3]=0 －> 2[(x-1)2+2]=0 -> (x-1)2+2 = 0

2[(x-1)2 + 2] = 0