(c-a)x^2 - 2(a-b)x + (b-c)=0

This question simply asks your knowledge about THE QUADRATIC FORMULA.

(c-a)x^2 - 2(a-b)x + (b-c) = 0

Therefore, THE DISCRIMINANT
= 4(a-b)^2 - 4(c-a)(b-c)
= 4a^2 - 8ab + 4b^2 - 4(bc - c^2 - ab + ac)
= 4a^2 - 4ab + 4b^2 - 4bc + 4c^2 - 4ac
= 4a(a-b) + 4b(b-c) + 4c(c-a)
= 4( a^2 - ab + b^2 - bc + c^2 - ca)
= 4[ (a-b)^2 + ab - bc + (c-a)^2 + ca - a^2 ]
= 4[ (a-b)^2 + (a-c)(b-a) + (a-c)^2 ]
> 0 <----- You can assum a > b > c for convenience

Since a, b, c are not all equal,
DISCRIMINANT does NOT EQUAL ZERO

Therefore, the equation has unequal real roots.

a=1, b=3,
DISCRIMINANT,
16 - 4(c-1)(3-c) > 0
16 + 4c^2 - 16c + 12 > 0
4c^2 - 16c + 28 > 0
c^2 - 4c + 7 > 0
(c-2)^2 > -3
(c-2)^2 > 0
c-2 > 0 OR c-2 < 0
c>2 or c< 2

2007-11-04 15:09:36 補充：
Hopefully it is right.