What is the connection between the discriminant of a quadratic and the distance formula?
Suppose that $a\ne 0$ and $b^2-4ac\geq 0$. Let $f(x)=ax^2+bx+c$ and $$ x_0= -\frac{b}{2a},\; x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, \;x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}. $$ It follows that $$ y_0=f(x_0)=\frac{-b^2+4ac}{4a},\;y_1=f(x_1)=0, \; y_2=f(x_2)=0. $$ The distances from vertex to roots: $$ d_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}=\sqrt{\left(-\frac{b}{2a}-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b^2+4ac}{4a}\right)^2}=\frac{\sqrt{(b^2-4ac)^2+4(b^2-4ac)}}{4|a|}>\frac{\sqrt{b^2-4ac}}{2|a|}, $$ $$ d_2=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2}=\sqrt{\left(-\frac{b}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b^2+4ac}{4a}\right)^2}=\frac{\sqrt{(b^2-4ac)^2+4(b^2-4ac)}}{4|a|}>\frac{\sqrt{b^2-4ac}}{2|a|}. $$
The connection you are looking for is simply Pythagoras theorem!
Imagine an upside down parabola. I ask that it be upside down so that the $y$ coordinate of the center is posiive. I do this just for convinience, my arguments are centainly true for any parabolas with two zeros.
You ask for a relationship between the equation for distance and the quadratic formula. Imagine two lines:
1) One connecting the center and the $x$-axis
2) One connecting the center and the zero.
The thing you have to see here, is that these two lines, when combined with the horizontal distance between the center and the zero, form a right triangle!
The length of line $1$ is simply the $y$ coordinate of the center, while the length of line $2$ (call it $H$) is given by the distance formula: $$H = \sqrt{(x_c-x_0)^2 + (y_c-y_0)^2}$$
And the horizontal distance is connected to these two by Pythagoras: $$y_c^2 + D^2 = H^2 \to D = \sqrt{H^2-y_c^2}$$
Your squinting was almost right. You just forgot that Pythagoras also gives these funny combinations of squares and square roots.
Let's do some algebra: The coordinates of the center are: $$x_c = \frac{-b}{2a} \quad \quad y_c = \frac{b^2(1-2a) + 2ac}{4a}$$
If we take the quadratic formula for granted, the coordinates of a zero are:
$$x_0 = \frac{-b}{2a} + \frac{\sqrt{b^2-4ac}}{2a} \quad \quad y_0 =0$$
We can calculate $H$, but since I'm lazy I'll calculate $H^2$: $$H^2 = (x_c-x_0)^2 + (y_c-y_0)^2$$ $$H^2 = \frac{b^2-4ac}{4a^2} + y_c^2$$
Note I left $y_c$ as is. You'll see why in a second:
The distance you want is:
$$D = \sqrt{H^2-y_c^2} = \sqrt{\frac{b^2-4ac}{4a^2} + y_c^2 - y_c^2} = \frac{\sqrt{b^2-4ac}}{2a}$$
So the link you were missing is the right triangle formed by the center, the zero and the $x$-axis. Note also that the distance formulate is derived from Pythagoras, so in reality the only connection is between Pythagoras and the quadratic formula.
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!