Is there a general way to solve exactly a system of this shape (the $a_i$ are constants):
$$\begin{array}{cc}a_1x^2+a_2x+a_3y^2+a_4y+a_5=0\\ a_6xy+a_7x+a_8y+a_9=0 \end{array} $$
It comes from a geometrical problem: the first equation states two vectors have same length, and the second ones state that they are perpendicular.
It can be reduced to one equation of degree four with one variable, but is it the best we can do?
Assume your $a_1\neq0,\,a_6\neq0$. Dividing by these, you can have the slightly simpler system with two fewer constants,
$$x^2 + b_1 x + b_2 y^2 + b_3 y + b_4 = 0$$
$$x y + b_5 x + b_6 y + b_7 = 0$$
Eliminating $y$ results in a monic quartic of form,
$$x^4+ax^3+bx^2+cx+d=0\tag1$$
The solution of the general quartic is not that bad. It is just,
$$x_{1,2} = -\tfrac{1}{4}a+\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag2$$
$$x_{3,4} = -\tfrac{1}{4}a-\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u-\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag3$$
where,
$$u = \frac{a^2}{4} +\frac{-2b+v_1^{1/3}+v_2^{1/3}}{3}$$
and the $v_i$ are the two roots of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
P.S. However, one has to be take note that some older CAS have $(-n)^{1/3} = \text{complex}$, even for real $n$.