假设 2x^2-3x+1 = y
(2x^2-3x+1)^2 = 22x^2-33x+1
(2x^2-3x+1)^2 = 22x^2-33x+11 - 10
(2x^2-3x+1)^2 = 11(2x^2-3x+1) - 10
y^2 = 11y - 10
y^2 - 11y + 10 = 0
(y-10)(y-1) = 0
y = 10 或 1
所以
2x^2-3x+1 = 10 或 1
当
2x^2-3x+1 = 10
2x^2-3x-9 = 0
(x-3)(2x+3)=0
x = 3 或-3/2
当
2x^2-3x+1 = 1
2x^2 - 3x = 0
x(2x-3)=0
x=0或3/2
所以
x = 0 或 3/2 或 -3/2 或 3