Given are $$S_1 = x(a_1x+2h_1y)+b_1y^2 + c_1=0$$ $$S_2 = x(a_2x+2h_2y)+b_2y^2 + c_2=0$$ $$S_1=S_2$$ I want to solve for a positive $x$ and positive $y$. There is a method as explained at the very end in this source, here's the quote:
Each of these equations represents a conic section, and they intersect at four points. We wish to find the point of intersection that lies in the all-positive quadrant - i.e. with x and y both positive. Form the equation $c_2S_1−c_1S_2=0$. This is also a quadratic equation representing a conic section passing through the four points. Furthermore, it has no constant term, and it, therefore, represents the two straight lines that pass through the four points. The equation can be factorized into two linear terms, αβ−0, where α=0 and β=0 are the two straight lines. Choose the one with positive slope and solve it with $S_1=0$ or with $S_2=0$ (or with both, as a check against arithmetic mistakes) to find x and y.
The source has given values for $a_1=11.73$, $h_1=17.06$, $b_1=14.93$ and $c_1=-20$ in its example $2.9.1$ which eventually shows that $x=0.2529$ and $y=1.000$
However, I don't really understand the method explained such as the factorization in two linear terms. How is this solved?
So you've gotten it to the form $$Ax^2 +Bxy + Cy^2 = 0$$ (the exact forms of $A, B, C$ will only get in the way right now).
The text tells you to write the left hand side as a product of linear factors:
$$Ax^2 +Bxy + Cy^2 = (ax + by)(cx + dy)$$ Except note that if you divide the first factor by a constant and multiply the second factor by the same constant, you get another factorization for the same expression. To get rid of that amiguity, assume $c = 1$:
$$Ax^2 +Bxy + Cy^2 = (ax + by)(x + dy)\\ Ax^2 +Bxy + Cy^2 = ax^2 +(ad + b)xy + bdy^2$$ Since this should hold for all $x, y$, we have $$a = A\\ad+b = B\\bd=C$$ which solves to $$\frac {AC}b + b = B\\b^2 - Bb + AC = 0\\b = \frac{B \pm \sqrt{B^2 - 4AC}}2$$
For each value of $b$ you can factor $$Ax^2 +Bxy + Cy^2 = (Ax + by)(x + \frac Cby)=0$$ which gives either $Ax + by = 0$ or $bx + Cy = 0$. This gives four lines in total: $$y = -\frac {2A}{B+\sqrt{B^2 - 4AC}}x\\y = -\frac {2A}{B-\sqrt{B^2 - 4AC}}x\\y = -\frac{B + \sqrt{B^2 - 4AC}}{2C}x\\y = -\frac{B - \sqrt{B^2 - 4AC}}{2C}x$$
These are the four lines the site talks about. Since you want both $x$ and $y$ positive, you can throw away any for which the slope is negative (this will depend on the values of $b_1, b_2, c_1, c_2, h_1, h_2$). Though I've not checked it myself, the site is claiming that there is always one with a positive slope, call it $y = mx$. Plug that back into either $S_1$ or $S_2$ (if you did the arithmetic right, either equation will give the same solution). The result will be a quadratic equation in $x$ only, which you can solve.
Once you know $x$, you also know $y = mx$, which gives the final solution.
Edit: Added demonstration on how to calculate that the four lines actually do satisfy the equation:
Let $$\omega = 1 \pm \sqrt{1 - \dfrac{4AC}{B^2}}$$
which gives $$\dfrac{4AC}{B^2} = 1 - (\omega - 1)^2 = 2\omega - \omega^2$$
The lines have the form $$y = -\frac {2A}{B\omega}x,\quad y = -\frac {B\omega}{2C}x$$
If you substitute the first expression for $y$ in $Ax^2 + Bxy + Cy^2$, you get $$\begin{align}Ax^2 + Bxy + Cy^2 &= Ax^2 - \frac {2A}\omega x^2 + \frac {4A^2C}{B^2\omega^2}x^2\\ &= \dfrac{Ax^2}{\omega^2}\left[\omega^2 - 2\omega + \dfrac{4AC}{B^2}\right]\\ &= \dfrac{Ax^2}{\omega^2}\left[\omega^2 - 2\omega + (2\omega - \omega^2)\right]\\ &= 0\end{align}$$
Had I used $y = \frac {2A}{B\omega}x$ as you claim in the comments, I would have gotten $\left[\omega^2 + 2\omega + (2\omega - \omega^2)\right] = 4\omega$ instead.
A similar calculation works for $y = -\frac {B\omega}{2C}x$ as well.