Why doesn't the quadratic equation contain $2|a|$ in the denominator?

One could take the square root as $2|a|$ instead, which would lead to:

$$ x+\frac {b}{2a} = \pm{\frac {\sqrt{b^{2}-4ac}}{2|a|}} \quad\iff\quad x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} \tag{1} $$

However, given that $\,|a|\,$ is either $\,a\,$ or $\,-a\,$ it follows that $\,\pm|a|=\pm a\,$, so the formula simplifies to:

$$ x = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{2|a|}} = -\frac {b}{2a} \pm {\frac {\sqrt{b^{2}-4ac}}{\color{red}{2a}}} = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \tag{2} $$

$(1)\,$ and $\,(2)\,$ are entirely equivalent, but $\,(2)\,$ is more convenient to use.

When taking the square root we put a $\pm$ on the right hand side to account for the two roots, so it is unnecessary to strip off the sign of $a$, as we will put it back anyways.

The two square roots of $a^2$ are $a$ and $-a$, sometimes written together as $\pm a$.

For real numbers $\pm a$ is equivalent to $\pm |a|$ but this is not true for complex numbers. So putting the absolute value operation in would make the proof less general.

We could write the proof as

$$ \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}} $$

$$ \pm\left(x+{\frac {b}{2a}}\right)={\frac {\pm\sqrt{b^{2}-4ac}}{\pm2a}} $$

$$ x+{\frac {b}{2a}}={\frac {\pm\sqrt{b^{2}-4ac}}{2a}} $$

But generally it is considered sufficient to put in just a single $\pm$ from the start rather than putting in one for each square root and then removing the redundant ones.