2(x^2-2x+6)+3根[x^2-2x+6]=27
设t=根号[x^2-2x+6]>0
则原方程是:2t^2+3t-27=0
(2t+9)(t-3)=0
t=-9/2(不符,舍)
t=3
所以:3=根[ x^2-2x+6]
x^2-2x+6=9
x^2-2x-3=0
[x-3][x+1]=0
x1=3
x2=-1
2(x^2-2x+6)+3根[x^2-2x+6]=27
设t=根号[x^2-2x+6]>0
则原方程是:2t^2+3t-27=0
(2t+9)(t-3)=0
t=-9/2(不符,舍)
t=3
所以:3=根[ x^2-2x+6]
x^2-2x+6=9
x^2-2x-3=0
[x-3][x+1]=0
x1=3
x2=-1