2x²-4x+3√(x²-2x+6)=15
2(x²-2x+6)+3√(x²-2x+6)-27=0
令t=√(x²-2x+6)≥0
方程化为2t²+3t-27=0
(t-3)(2t+9)=0
解得t=3或t=-9/2(舍去)
∴√(x²-2x+6)=3
x²-2x+6=9
x²-2x-3=0
(x-3)(x+1)=0
解得x=3或x=-1
2x²-4x+3√(x²-2x+6)=15
2(x²-2x+6)+3√(x²-2x+6)-27=0
令t=√(x²-2x+6)≥0
方程化为2t²+3t-27=0
(t-3)(2t+9)=0
解得t=3或t=-9/2(舍去)
∴√(x²-2x+6)=3
x²-2x+6=9
x²-2x-3=0
(x-3)(x+1)=0
解得x=3或x=-1