One maths question

(x^2- 4x)^2 - 4(x^2 -4x) = 5
Let y = x^2 - 4x
y^2 - 4y - 5 = 0
(y-5)(y+1)=0
y = 5 or -1
x^2 - 4x = 5 or x^2 - 4x = -1
x^2 -4x-5 = 0 or x^2 - 4x+1 = 0
(x-5)(x+1) = 0 or x= [4+rt(12)]/2 or [4-rt(12)]/2
x = 5, -1, 2+rt(3) or 2-rt(3)

(x^2- 4x)^2 - 4(x^2 -4x) - 5 = 0
[(x^2 -4x) - 5] [(x^2 -4x) + 1] = 0
(x^2 -4x) = 5 or (x^2 -4x) = -1
x^2 -4x - 5 = 0 or x^2 -4x + 1 = 0
(x-5)(x + 1) =0 or x = 2 + sqrt(3) or x = 2 - sqrt(3)
x = 5 or x = -1 or x = 2 + sqrt(3) or x = 2 - sqrt(3)

2007-09-04 21:41:37 補充：
慢左, 不過... 請多多支持