f（x）=x^2+ax+b
A=｛x| x=f(x)｝, B={x| x=f(f(x))}
x ∈ A
=> x = f(x)
=> f(x) = f(f(x))
=> x = f(f(x))( x= f(x))
=> x∈ B
=> A is subset of B
A={-1,3}
x= x^+ax+b
x^2+(a-1)x+b=0
sum of roots
-(a-1) = 2
a = -1
product of roots
b=-3
=> f(x) = x^2-x-3
f(f(x)) = (x^2-x-3)^2 -(x^2-x-3) -3 = x
(x^2-x)^2-6(x^2-x) + 9 - x^2=0
x^4-2x^3-6x^2+6x+9=0
(x+1)(x-3)(x^2-3)=0
x = -1 or 3 or -√3 or √3
B = {-1 , 3 , -√3 , √3}