<s>[INST] 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?
 [/INST] Without loss of generality, we can find the horizontal distance between $x=\dfrac{-b}{2a}$ and $x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}.$  So inputting these coordinates into the distance formula, with both $y-$coordinates equal to zero, we have
\begin{align}
d&=\sqrt{\left(\dfrac{-b+\sqrt{b^2-4ac}}{2a}-\dfrac{-b}{2a}\right)^2+(0-0)^2}\\
&=\dfrac{\sqrt{b^2-4ac}}{2a}
\end{align}
This is the horizontal distance from the positive root of $f(x)=ax^2+bx+c$ "to the right" of the point $\left(\dfrac{-b}{2a},0\right).$  So both roots are located this distance left and right of the vertex's $x-$coordinate, giving the roots $x-$coordinates
$$\left(\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},0\right).$$

As a side note, the quadratic formula is derived from completing the square as follows:
\begin{align}
y&=ax^2+bx+c\\
&=a\left(x^2+\frac ba x+\frac ca\right)\\
&=a\left(x^2+\frac ba x+\frac{b^2}{4a^2}+\frac ca-\frac{b^2}{4a^2}\right)\\
&=a\left(x+\frac b{2a}\right)^2+c-\frac{b^2}{4a}.
\end{align}
Now letting $y=0,$ we get
\begin{align}
a\left(x+\frac b{2a}\right)^2&=\frac{b^2}{4a}-c\\
\left(x+\frac b{2a}\right)^2&=\frac{b^2}{4a^2}-\frac ca\\
\left(x+\frac b{2a}\right)&=\pm\sqrt{\frac{b^2}{4a^2}-\frac ca}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\
x&=\dfrac{-b+\sqrt{b^2-4ac}}{2a}.
\end{align}
</s>