For $c<1$, let $f_c(x)=x^2+c$, determine the period-2 points? Ok so I know I need to set $$f_c^2(x)=x$$ so:
$$(x_0^2+c)^2+c = x \iff x_0^4+2cx_0^2-x_0+c^2+c=0$$
But how do I then solve this? Ok I have solved for the four set of roots, now you can see there are two roots for $c<\frac{1}{4}$ and four for $c<-\frac{3}{4}$. How do I find the hyperbolicity and therefore the nature of these points?

As user user58697 pointed out $x_0^2 + c - x_0$ and $x_0^4+2cx_0^2-x_0+c^2+c$ are not coprime.
Indeed, $x_0$ is a fix point of $f$ $\iff f(x_0)=x_0\iff x_0^2+c-x_0=0$. 
Since fix points of $f$ satisfy 
$$
f(f(x_0))=x_0
$$
as well,
then fix points are also solutions of 
$$
x_0^4+2cx_0^2-x_0+c^2+c=0.
$$
Hence, in this case you may divide $x_0^4+2cx_0^2-x_0+c^2+c$ over $x_0^2 -x_0+ c$ in order to get the polynomial $g(x)$ of (proper) period-2 points of $f$. In this case $g(x)$ is of degree two, so you may solve it.
