number of solutions of $f(f(f(f(x))))$ Let $f(x)={x^2+10x+20}$ then the number of real solutions of $f(f(f(f(x))))=0$
My try: I first tried finding the roots of $f(x)=0$ they came out to be
$-5+\sqrt{5}$ and $-5-\sqrt{5}$.
Let these roots be $\alpha$ and $\beta$
thus $f(\alpha)=0$ also if $f(f(f(f(x))))=0$ then $f(f(f(x)))=\alpha$
I was able to go only till this. I skipped the method of actually finding the composite function as it would be too cumbersome. Any ideas, helps will be appreciated.

The multiple composition of the equation can also be viewed in this way (amounting to a different interpretation of what is described in many of the other posted answers).  The function $ \ f(x) \ = \ x^2 + 10x + 20 \ = \ (x + 5)^2 - 5 \ \ $ has the two real zeroes $ \ r_{\pm} \ = \ -5 \ \pm \sqrt{5} \ \ , $ so the next level of composition, $ \ f( \ f(x) \ ) \ = \ 0 \ \ $ requires us to solve $ \ f(x) \ = \ -5 \ \pm \ \sqrt{5} \ \ ; $ further levels of composition seem to call for more daunting calculation.
We might instead look at how the function transforms numbers.  The fixed points of the function are given by
$$ f(x) \ = \  x \ \ \Rightarrow \ \ x^2 \ + \ 10x \ + \  20 \ \ = \ \ x \ \ \Rightarrow \ \ x^2 \ + \ 9x \ + \  20 \ \ = \ \ (x + 5)·(x + 4) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ x \ \ = \ \ -4 \ , \ -5 \ \ . $$
So $ \ f(-4) \ = \ -4 \ \ $ and $ \ f(-5) \ = \ -5 \ \ . $  We could then ask about how other numbers are "mapped" by the function relative to the fixed points.  For $ \ x \ = \ -5 + \alpha \ \ , $ we obtain
$$   ( \ [-5 + \alpha] + 5)^2 - 5 \ \ = \ \ \alpha^2 \ - \ 5 \ \ , $$
We see from this that $ \ f(x) \ = \ 0 \ \ $ corresponds to the mapping of $ \ -5 + \alpha \ \rightarrow \ 0 \ \ $ such that $ \ \alpha^2 - 5 \ = \ 0 \ \ . $  Hence there are two such points, $ \ -5 \pm \sqrt5 \ \ , $ which are symmetrically located around the axis of symmetry of the parabola representing the quadratic polynomial (as we expect of the zeroes).  What we now also see is that $ \ f( \ f(x) \ ) \ = \ 0 \ $ corresponds to $ \ -5 + \alpha' \ \rightarrow \ -5 + \sqrt5 \ \ $ such that $ \ (\alpha')^2 - 5 \ = \ ( \ \alpha^2 \ )^2 - 5 \ = \ 0 \ \ . $
Repeated applications of the function transformation maintain the symmetry of the two points about $ \ x = -5 \ \ $ while "compressing" them successively closer to this fixed point (which would be referred to as an "attractor").  As also described in comments and the other posted answers, we observe that $ \ f^3( x ) \ = \ 0 \ $ is solved by  $ \ -5 + \alpha'' \ \rightarrow \ -5 + \alpha' \ \ $ with $ \ (\alpha'')^2 - 5 \ = \ ( \ ( \ \alpha^2 \ )^2 \ )  ^2 - 5 \ = \ 0 \ \  $  and finally $ \ f^4( x ) \ = \ 0 \ $ is solved by  $ \ -5 + \alpha'''  \ \ $ with $ \ (\alpha''')^2 - 5 \ = \ ( \ ( \ ( \ ( \ \alpha^2\ )^2 \ )^2 \ )  ^2 - 5 \ = \ 0 \ \Rightarrow \ \alpha''' \ = \ \pm 5^{1/16} \ \ . $  At any finite number of compositions then, we continue to obtain just two real zeroes symmetrically placed about $ \ -5 \ $ (and a whole slew of complex zeroes).
What about the other fixed point?  If we use $ \ x \ = \ -4 + \beta \ \ , $ we obtain
$$   ( \ [-4 + \beta] + 5)^2 - 5 \ \ = \ \ ( \ 1 + \beta \ )^2 \ - \ 5 \ \ = \ \ 0 \ \ \Rightarrow \ \ \beta \ \ = \ \ -1 \pm \sqrt5 \ \ , $$
leading us again to $ \ x \ = \ -5 \pm   \sqrt5 \ \ . $  (So the fixed point $ \ -4 \ $ acts as a "repellor".)
