Approximation of the quadratic formula with straightedge and compass Given a directrix and a focus (blue), we can define a parabola as illustrated below. We suppose the parabola intersecting the $x$-axis in correspondence of the red dots.

We draw the line perpendicular to the $x$-axis and passing through the focus. This line intersects the $x$-axis in the point $H$ and the directrix in the point $I$. By means of these two points, we can draw two circles: one with center in $H$ and passing through the focus (green), and one with center in  $H$ and passing through $I$ (brown).

The directrix intersects the $y$-axis in correspondence of the point $J$. We draw now the tangent to the green circle passing through $J$, obtaining the point $K$ on the brown circle.


If the roots are located inside the green circle, then the circle passing through $(0,0)$, $K$ and $J$ (red) intersects the $x$-axis in a point which lies in close proximity to one of the roots of the parabola.



I tried to find a simple way to connect this construction with the quadratic formula, but I didn't go far. A little help would be greatly appreciated. 

Moreover, are you aware of a better geometric construction of the quadratic formula? Is it possible to produce an exact construction with straightedge and compass?
I apologize for the naivety. Thank you very much for your help!

The construction is in fact trivial. The parabola being the locus of points equidistant to the focus and the directrix, you draw a circle around the focus, with radius equal to the distance between the directrix and axis.

From the geometric definition, the equation of the parabola is
$$(y-y_d)^2=(x-x_f)^2+(y-y_f)^2$$ and you are looking for the points such that $y=0$, or
$$y_d^2=(x-x_f)^2+y_f^2.$$
The solution is obviously
$$x=x_f\pm\sqrt{y_d^2-y_f^2}$$ where you recognize pieces of the quadratic formula.
