What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the center
$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula
$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$
Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?

Consider the image below.

The roots of $f(x)=ax^2+bx+c$ will lie on the circle in the complex plane whose center is $\dfrac{-b}{2a}.$  In the case where both roots are real numbers, they will lie on the real line at $R_1$ and $R_2.$  In the case where the roots are purely imaginary, they will lie on the vertical axis at $ir_1$ and $-ir_1.$  Now we want to find $d$ the radius of the circle.
To do this, note that the product of the roots of $f(x)=ax^2+bx+c$ is $\dfrac ca.$  So in the case that the roots are purely imaginary, we have
$$\frac ca=(ir_1)(-ir_1)=r_1^2,$$
And thus $r_1=\sqrt{\dfrac ca}.$  Note these roots will only occur when $b=0$ and $ac>0.$
Now to find the distance $d,$ we use the Pythagorean Theorem to get
$$d^2=\left(\frac{-b}{2a}\right)^2+\left(\pm i\sqrt{\frac{c}{a}}\right)^2=\frac{b^2}{4a^2}+\frac{-c}{a}=\frac{b^2-4ac}{4a^2},$$
and thus
$$d=\frac{\pm\sqrt{b^2-4ac}}{2a}.$$
In the case where the roots are both real, for $b^2>4ac$ we have
$$R_1=\frac{-b-\sqrt{b^2-4ac}}{2a},\quad\text{and}\quad R_2=\frac{-b+\sqrt{b^2-4ac}}{2a}.$$

One may struggle or take issue about the above argument, since $d$ is not a distance as traditionally defined in the complex plane.  See related question:
Pythagorean Theorem for imaginary numbers
It should be noted that my notion of distance is being done on the  hyperbolic space.  In both the Euclidean plane and the Hyperbolic plane, the distance between points on the real line is the same.  But once we venture off the real line we get different versions of the Pythagorean Theorem.
In Euclidean space, we have $\cos^2t+\sin^2t=1\Longrightarrow a^2+b^2=c^2.$
In Hyperbolic space, we have $\cosh^2t{\color{red}-}\sinh^2t=1\Longrightarrow a^2-b^2=c^2\equiv a^2+(\pm ib)^2=c^2.$
You may ask:

Why do we need to represent our roots on the hyperbolic plane?

Answer:

Because it allows us to use imaginary numbers as lengths in the Euclidean plane!

