Geometric derivation of the quadratic equation The quadratic equation can be thought of as specifying distances in the Euclidean plane. It tells us that the $x$-intercepts of a function occur at a distance of $\frac{\sqrt{b^2-4ac}}{2a}$ from the $x$ coordinate of the max/min point. Does anyone know a purely geometric derivation of the quadratic formula related to this fact?

Problem:
Let $V(x_V,y_V)$ the vertex of a parable whose leading coefficient is $a$. Find out its roots through geometric construction.
Resolution:


*

*Draw line $e$ perpendicular to $x$-axis, such that $V \in e$.

*Let $A=e\cap x$-axis, mark points $B$ and $C$ on $x$-axis, such that $AB=|a|$ and $AC=1$.

*Mark point $F$ on $e$, such that $FA=1$, and $V$ and $F$ are in different semi-planes relative to $x$-axis.

*Draw a line $j$ parallel to $VB$ through $C$.

*Let $E=j \cap e$. Mark point $G$ on $e$, such that $G$ is the midpoint of $EF$.

*Draw a circle $\lambda$ centered at $G$ and whose radius is $GF$.

*Let $\{H,I\}=\lambda \cap x$-axis. The abscissas of $H$ and $I$ are the roots of the quadratic equation that we get when the parabola crosses the $x$-axis.


The picture below shows an example.

Edit/Justification
Recall that:
$$x_V= -\frac{b}{2a} \quad (1)$$
and
$$y_V=-\frac{\Delta}{4a}. \quad (2)$$
Note that:
$$x_A=x_V\quad (3)$$
and
$$AV=|y_V|.\quad (4)$$
Note also that $\triangle ABV \sim \triangle ACE$, hence:
$$\frac{AB}{AV}=\frac{AC}{AE}. \quad (5)$$
From $(2)$, $(4)$ and $(5)$ we get:
$$AE=\frac{|\Delta|}{4a^2}. \quad (6)$$
Recall that an inscribed triangle whose major side is a diameter is right-angled. Therefore $\triangle FHE$ and $\triangle FIE$ are right-angled at $H$ and $I$ respectively.
Note that $AH=AI$, and that $AI$ is a height of $\triangle FIE$.
As $\triangle FIE$ is a right-angled triangle we have:
$$AI^2=AF \cdot AE. \quad (7)$$
Recall that by construction $AF =1$. 
So we get from $(6)$ and $(7)$:
$$AI=AH=\frac{\sqrt{|\Delta|}}{2a}. \quad (8)$$
Note that:
$$x_I=x_A+AI \quad(9)$$
and
$$x_H=x_A-AH. \quad(10)$$
From $(1)$, $(3)$, $(8)$, $(9)$ and $(10)$, we get:
$$x_I=-\frac{b}{2a}+\frac{\sqrt{|\Delta|}}{2a} \quad (11)$$
and
$$x_H=-\frac{b}{2a}-\frac{\sqrt{|\Delta|}}{2a} .\quad (12)$$
