This is a question from On Functions and Functional Equations by J. Smital, specifically Exercise 3.2. I feel like I must be missing something here, or some additional context is inferred which I am not picking up on.
The question asks us to (verbatim),
Find a polynomial function of degree 2 having two fixed points, both of them being computable by the method of iterations starting in some suitable neighbourhood of the fixed point.
The assumption here is that we are working in $\mathbb{R}$. Now this would seem to imply both fixed points are attractive or, at minimum, not repulsive. However if we start with a general quadratic $f(x) = a x^2 + b x + c$ then, without going into too much detail, the fixed points can be computed using the quadratic formula as,$$x_1, x_2 = \frac{- b + 1 \pm \sqrt{(b - 1)^2 - 4 a c}}{2 a}.$$
From which it follows that, $$f'(x_1), f'(x_2) = 1 \pm \sqrt{(b-1)^2 - 4 a c}.$$
Which tells us that either $|f'(x)| < 1$ for one fixed point and $|f'(x)| > 1$ for the other, so we have an attractive/repelling pair. Or $|f'(x)| > 1$ for both fixed points so we have a dual repelling pair. At least according to Theorem 3.4. of the same book. That is, unless $(b-1)^2 = 4 a c$, in which case we only have a single fixed point at all.
I cannot see any way of solving this problem, at least whilst remaining restricted to the reals. However everything in the book so far has been in $\mathbb{R}$. All the theorems presented are defined in terms of intervals on the real real. Perhaps I am misunderstanding the question here?