Two quadratic equations, more than two solutions? It's clear that a system of two quadratic equations can have none, one or two solutions. 
For example: $y = x^2 + 2$ and $y = - x^2 + 1$ have none. $y = x^2$, $2x^2 - 8x + 8$ and $y = - x^2 + 8x - 8$ have $4$ as common solution. And $2x^2 - 8x + 8 = x^2 - 4x + 5$ have $1$ and $3$ as solutions.
Is it also possible to have more solutions? Intuitively I'd say that two is the max number of solutions, but where is the proof of this?
Note (I found it after asking the question, and receiving answers): the question On the number of possible solutions for a quadratic equation. is strongly related to this question.
 A: It depends what you mean by a system of quadratics.
If you have $y=ax^2+by+c$ and $y=px^2+qx+r$ then you must have $(a-p)x^2+(b-q)x+(c-r)=0$. This equation has at most two solutions for $x$ (unless $a=p, b=q, c=r$) and each solution for $x$ gives a value for $y$.
However there is another way of thinking about quadratic curves in the plane, which would include circles (like x^2+y^2=r^2), hyperbolae (for example, xy=c), and ellipses - and rotated versions too. A parabola can cut a circle in four points, for example.
A: Intuitively you are correct however you have forgotten a trivial case, when both quadratics are the same in which case you have infinitely many solutions.
Apart from this very specific case here's how to prove that indeed they only have none,one or two solutions.
Consider a system where you have :
\begin{cases}
y &=& a_1x^2 + b_1x + c_1\\
y &=& a_2x^2 + b_2x + c_2
\end{cases}
Since $y$ equates both second order polynomial, you can equate them and obtain the following equation :
$$a_1x^2 + b_1x + c_1 = a_2x^2 + b_2x + c_2$$
This clearly has at most two solutions of $x$. Substitute back all solutions and this will give you a corresponding $y$ for each solution of $x$.
