$f(f(x))$ has no  fixed points if $f(x)$ has no fixed points 
Assume that $f(x)=x$ has no real roots where 
$$f(x) = ax^2+bx+c$$   Prove that $f(f(x))=x$ has no real roots as
  well.

What I've done is, calculating $f(f(x))$:
$$f(f(x))=a(ax^2+bx+c)^2+b(ax^2+bx+c)+c$$
and putting $\Delta=b^2-4ac<0$ which seems quite time consuming. Is that the right thing to do?

Hint:
$f(x) - x$ is always of the same sign...
To use the hint:
If $f(x) \gt x$ for all $x$, then $f(f(x)) \gt f(x) \gt x$. 
Similarly, if $f(x) \lt x$ we can show that $f(f(x)) \lt x$.
(For more details, see Arturo's answer).
Since this is tagged algebra precalculus, here is a continuity free proof to show that $f(x) - x$ is of the same sign.
We will first show that, if $g(x) = px^2 + qx + r$, $p \gt 0$,  is a quadratic and if there is a real number $s$ such that $g(s) \lt 0$ then $g(x) = 0$ has a real root.
By completing the square, we have 
$$g(s) = p\left(s + \frac{q}{2p}\right)^2 + r - \frac{q^2}{4p} \lt 0$$
i.e.
$$0 \le p\left(s + \frac{q}{2p}\right)^2 \lt -r + \frac{q^2}{4p}$$
Thus $q^2 \gt 4pr$ and $g(x) = 0$ has a real root.
The case $p=0$ is easy to deal with.
Now let $s,t$ be such that $f(s) \lt s$ and $f(t) \gt t$. If $a \ge 0$, we can apply the above to $g(x) = f(x) - x$, else, we apply it to $g(x) = x - f(x)$
