[Scope of level: Undergrad Abstract Algebra]
This was a bonus topic we covered to very little extent so I'm rather unsure on how you go about doing this. From the notes we were given it considers the quadrics to be degenerates which then you can express them as factors of linear polynomials then solve separately.
If the quadric $P = Q = 0$ (of the form $ax^2 + 2bxy + cy^2 + 2dx +2ey + f$) are not degenerate then you construct an equivalent system that is degenerate by taking $(P') + k (Q')$ and find $k$ such that $P = P + k Q = 0$ where $(P')$ is a symmetric matrix constructed by taking the coefficients of the quadric $((P') = [(a,b,d),(b,c,e),(d, e, f)]$ by taking $det((P') + k(Q'))=0$.
(If need be, can provide link to said notes (in .pdf))
For example, consider the following two quadrics:
$2x^2 - xy + 3y^2 = 36$
$3x^2 - 4xy + 5y^2 = 36$
I tried the above steps and the constant required turns out to be some nasty number which really complicates computations (by quadratic formula as it forms a cubic polynomial on $k$ but with $(-36-36k)$ already factored out leaving a quadratic portion to be solved) which sends me the message that this is not right. So I just tried taking the difference of the two quadrics (which is essentially the case of $k=-1$) and solved directly as so and got $\{(4,2), (-4,-2), (3,3), (-3,-3)\}$ as my intersection points.
Is such approach acceptable or am I not seeing the picture here? Since I did not consider other possible values of $k$ I feel as if this misses out on some subset of solutions.
Any advice to point me in the right direction would be greatly appreciated :)