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In order to find the solutions of the system of conics you mention, it suffices to give a procedure to find the projection of the simultaneous vanishing set of the two conics onto some axis, for instance, the $y$-axis: the coordinates of the projection onto the $y$-axis are the $y$-coordinates of the intersection points. Knowing them allows you to substitute back into the equations the values of $y$ and solve a system of equations in a single variable $x$, thus making the problem simpler. In terms of ideals, suppose that we can find a non-zero element $r$ in the ideal generated by the two conics, that depends only on a single variable, say $y$. This means that every solution of the system has $y$-coordinate satisfying the polynomial $r$. Thus we have severely limited the choices for the $y$-coordinates of the intersection (namely, they all have to satisfy the polynomial $r$) and, provided we can actually solve the polynomial $r$ in one variable, we can then substitute the various values of $y$ we found back into the initial equations and solve for $x$, again using our algorithm for solving polynomials in a single variable. So far, so good, I hope! The question is how to produce the element $r$. This is where Gröbner bases come in.

In order to do the computation you ask, one would have to choose an appropriate monomial order. I will gloss over this, and simply "do the obvious". It is an exercise for you to figure out which order to use so that the computation I am going to make is actually a Gröbner bases computation. Scattered throughout the computation there will be also a couple of special cases that I will not deal with: again, you can treat those as exercises for you!

First, I am going to choose a generic basis, so that the first conic has the form
$$
x^2 + alpha x + beta ,
$$
where $alpha$ and $beta$ are polynomials in $y$ only (I am trying to simplify the notation as much as possible; the only "assumption that I have made is that the coefficient of $x^2$ is non-zero, which can be arranged unless the "conic" is defined by a polynomial of degree at most one).

In this basis, the second equation can be assumed to have no $x^2$ term (indeed, eliminating the $x^2$ term using the first equation would be the first step in any reasonable Gröbner basis computation in which the term $x^2$ is the highest term in sight). Thus I am going to write the second conic as
$$
gamma x + delta
$$
where, as before, $gamma$ and $delta$ are polynomials in $y$ only. Again, just to fix on a definite case, I am going to assume that the polynomial $gamma$ is non-zero. (If $gamma$ were zero, then we would have found a polynomial in the ideal generated by the two conics which only depends on $y$: this was our goal at the start anyway! Obviously, you should worry about the case $gamma=delta=0$, but I won't.)

To compute a Gröbner basis, you would now compute S-polynomials: let's do it here as well. I am going to try to eliminate the $x^2$ term from the first equation, by using the second equation. This is easy: multiply the first equation by $gamma$ and the second one by $x$ and take the difference: we are left with the equation
$$
(alpha gamma - delta) x + beta gamma .
$$
Now we are going to eliminate the $x$ term from this last equation using again the second equation: multiply the second equation by $(alpha gamma - delta)$, the last equation by $gamma$ and subtract to obtain
$$
delta^2 - alpha gamma delta + beta gamma^2 .
$$
We found an expression independent of $x$!! We are done... provided this expression is not identically zero. You can figure out what this would mean and what happens in this case. Note also that the final expression is what we would have obtained if, at the very beginning, we had "solved" $x=-frac{delta}{gamma}$ using the second equation, substituted in the first equation and cleared the denominators.

I hope that this "hybrid" computation explains what is going on: Gröbner bases and Buchberger's algorithm are a systematic way of "solving" systems of equations. You do not have to do any thinking, once you set up the problem. But you need to set up the problem so that it computes what you want. In this case, you could have used several shortcuts to get to the answer, without following all the steps. In more complicated situations, Buchberger's algorithm might be the best way of keeping track of all the steps to be taken.

Let me also comment that, except in the case in which you have a computer doing the computations for you, it is highly unlikely that Gröbner bases will help you with a specific question, unless you could have also found simple tricks to solve it right away.

You have to read through the book [David A. Cox, John B. Little and Don O'Shea, Ideals, Varieties, Algorithms].

In any case, if you start with a system of polynomial equations and compute a Groebner basis for the ideal they generate, you get a "maximally triangular" system of equations which is equivalent to the original one---that is why Groebner bases generalize Gaussian elimination.

Let me do a simple example using Macaulay 2. Consider the ring $mathbb Q[x,y,z]$:

i1 : R = QQ[x, y, z, MonomialOrder => Lex]



o1 = R



o1 : PolynomialRing

the Fermat quintic surface $x^5+y^5+z^5=1$, whose ideal is

i2 : surface = ideal (x^5 + y^5 + z^5 - 1)



             5    5    5

o2 = ideal(x  + y  + z  - 1)



o2 : Ideal of R

and the curve $x^2=y^2$, $y^3+1=z^3$:

i3 : curve = ideal (x^2 - y^2, z^3 - y^3 - 1)



             2    2     3    3

o3 = ideal (x  - y , - y  + z  - 1)



o3 : Ideal of R

The ideal of the intersection of the surface and the curve is the sum of the ideals of the intersecands:

i4 : intersection = curve + surface



              2    2     3    3       5    5    5

o4 = ideal (x  - y , - y  + z  - 1, x  + y  + z  - 1)



o4 : Ideal of R

The number of points on the intersection, counting multiplicities, is the degree of the ideal:

i5 : degree intersection



o5 = 30

if we now compute the degree of the radical of the ideal, we get a lower number, so not all the points are simple:

i6 : degree radical intersection



o6 = 25

Now look at the lexicographic Groebner basis of the ideal of the intersection:

i7 : ideal gens gb intersection



              17      16      15      14      13      12       11       10      9      8       7       6      5      4      3  

o7 = ideal (9z   + 27z   + 54z   + 50z   + 15z   - 63z   - 104z   - 108z   - 35z  + 35z  + 108z  + 104z  + 63z  - 15z  - 50z  -

     ----------------------------------------------------------------------------------------------------------------------------

        2                 5            16      15      14      13      12       11      10      9       8      7       6       5

     54z  - 27z - 9, 20y*z  - 20y + 27z   + 18z   + 45z   - 75z   - 35z   - 164z   + 79z   - 10z  + 230z  - 10z  + 119z  - 164z 

     ----------------------------------------------------------------------------------------------------------------------------

          4       3      2              3    3                                       16        15        14        13        12  

     - 35z  - 155z  + 45z  + 18z + 67, y  - z  + 1, 50x*z - 50x + 50y*z - 50y + 1242z   + 2718z   + 4770z   + 2220z   - 1235z   -

     ----------------------------------------------------------------------------------------------------------------------------

          11        10        9        8        7        6        5        4        3        2                 2    2

     7979z   - 7481z   - 6010z  + 1810z  + 5240z  + 9394z  + 5346z  + 1240z  - 4030z  - 4005z  - 2657z - 583, x  - y )



o7 : Ideal of R

The first generator depends only on $z$. The second one and the third, only on $z$ and $y$, and the fourth depends (linearly) on $x$ too. This is a "triangular" system of equations, from which you can compute the 30 points of intersection, assuming you get to solve the first equation for $z$, which is of degree 17...

The example that you propose is a bit too general to serve as an introductory pedagogical example (compare e.g. the famous Kahan SIGSAM problem 9 of determining conditions for an ellipse to lie inside the unit circle). Instead, begin with one of the earliest historical examples: Gauss's proof that every symmetric polynomial can be written uniquely as a polynomial in the elementary symmetric polynomials. This is one of the first and the simplest applications of rewriting reduction via a lexicographic order. For a nice presentation see Chapter 7 of Cox, Little, O'Shea: Ideals, Varieties and Algorithms. They also present generalizations to the ring of invariants of a finite matrix group $rm G subset GL(n,k)$. You can find online copies of said book via various ebook databases.

You might find Lecture 3 of James Carlson's CIMAT Lectures of value as well as other material on this page:

http://www.math.utah.edu/~carlson/cimat/

This is a simplistic version of the Gröbner basis computation and complements Mariano's answer. The idea behind the Buchberger algorithm is to build a basis so that you are to be able to do "unique division" between nonlinear polynomials. This is done by computing (what is called the) S-polynomial between any 2 given polynomials until there are no more produced.

Decide some ordering (grlex, lex, revlex)

Initialize I = <f1, f2> 

where we use the polynomials you gave above.

f_1=ax^2+2bxy+cy^2+dx+fy+g

f_2=ax^2+2bxy+cy^2+dx+fy+g

Set n=2



Repeat the following in a loop

    Iteratively choose any two polynomials ( f_i, f_j ) from set I 

    n = n + 1

    Compute f[n] = S-polynomial( f_i, f_j )

Until all S-polynomials return 0

At the end of the computation, you will have a set I = {f1,f2, ..., fk} such that one of the polynomials will be in only one term (say 'x'). Solve that equation to get the one coordinate of the intersection, and then work upwards through the rest of the polynomials.

The general question is related to Elimination theory, and AFAIK there are three basic methods ( I can't speak of their complexity )

1. Gröbner basis, as seen above


2. Use Resultants (example below)


3. Wu's method of characteristic sets, which I can't say much about.

If you wanted to use resultants to find the intersection of say

E={f1(x, y,z), f2(x, y, z), f3(x, y, z)},

you would first compute

E1 = {g1 = Resultant_in_x (f1, f2), g2 = Resultant_in_x(f2, f3)} 

and then

E2 = { h1 = Resultant_in_y(g1, g2} }

The last polynomial h1 would be in one variable and amenable to solution.