$f(f(x))$ 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?

Write $g(x)=f(x)-x$.  Then $g$ is continuous and never zero, so it must be either always positive or always negative.
Now $f(f(x))-x = g(f(x))+g(x)$, which is always positive if $g$ is always positive, or always negative if $g$ is always negative.  In either case, it's never zero, so $f(f(x))$ is never equal $x$.
