how can I find roots of $x^2+px+q=0$ using iterative methods? I guess there are two real roots as long as $p^2 \gt 4q$
In order to use Newton's method, $N(x)=\cfrac{x^2-q}{2x+p}$
What should be the starting value and how can I guarantee that by iterating this Newton's function $p^2 \gt 4q$? 
Or is there any other way to find roots of this function using other iterative methods other than Newton's? 
 A: So your question is satisfactorily answered if you are given initial conditions for Newton's method which find each root. Here I will assume that two roots exist, i.e. $p^2-4q>0$.
Given an initial condition not exactly on the vertex, the Newton iteration will stay on that side of the vertex, because of the fact that there is only one turning point. Also, because of global concavity/convexity, the sign of the quadratic will alternate between iterates. With this idea and a bit of brute force algebra, I think one can prove that the sequences of double iterates, i.e. $x_0,x_2,\dots$ and $x_1,x_3,\dots$ both converge monotonically to the root.
Thus I think we have that if $x_0>-p/2$ then the sequence converges to the larger root and if $x_0<-p/2$ then the sequence converges to the smaller root.
Knowing something about the roots you can make better guesses. For instance you could start with a linear approximation of the square root in the quadratic formula, which gives approximate roots of
$$\frac{-p \pm \left ( p-\frac{4q}{2p} \right )}{2} = -\frac{q}{p},-p+\frac{q}{p}.$$
This works well when $p^2>4|q|$, preferably by quite a bit. If $p^2<4|q|$, then $\pm \sqrt{-q}$ will definitely perform better.
A: Either use the quadratic formula (since your equation is quadratic) or let  $f(x)=x^2+px+q=0$, 
then $$f^{\prime}(x)=2x+p$$ and by Newton-Rapson method $$x_{n+1}=x_n-\cfrac{f(x_n)}{f^{\prime}(x_n)}$$ Starting with $x_0=1$ this will be $$x_{n+1}=x_0-\cfrac{x^2+px+q}{2x+p}=x_{n+1}=1-\cfrac{1+p+q}{2+p}$$ Normally, when we want to chose a starting value we seek an integer such that $f(x)$ is as close to zero as possible which gives the function the best possible chance of converging. 
So $$f(1)= 1 + p + q\tag{1}$$ and $$f(-1)= 1 - p - q\tag{2}$$ Adding $(1)$ to $(2)$ and taking the average gives $1$. So we start with $x_0=1$.
