Each equation describes a conic. Thus, you are trying to compute the intersection $\cal C_1\cap\cal C_2$ of two conics. In general, you have to expect four intersection points (e.g., think of two ellipses meeting transversally).
There are special cases where the task is simplified. For instance, if one of the two conics splits as the union of two lines, or if one of the equations reduces easily to the form $y=f(x)$.
In the general case you can either try Elimination Theory or you can exploit the fact that conics are rational curves, i.e. that there is a parametrization
\Bbb R\ni t\mapsto (x(t),y(t))\in\cal C_1\qquad(*)
given by rational functions, i.e. quotient of polynomials in $t$. Then if you plug this "rational description" of $\cal C_1$ into the equation of $\cal C_2$ will eventually get a polynomial equation of degree 4 in the variable $t$ only, whose solutions correspond to the intersection points. In order to solve this degree 4 equation you need either some luck, or some patience to go browsing old Algebra books.
In order to get (*), the usual method is to find just one point $(x_o,y_o)\in\cal C_1$, consider the lines
through it and describe the second intersection of $\cal C_1\cap\ell_t$ in terms of $t$.