Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$? I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
However, this strikes me as not being simplified. Isn't it more proper to write it like this?
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$
$$x=\frac{-b\pm((b-2)\sqrt{ac})}{2a}$$
$$x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$$
Why isn't $x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$ more commonly used as the quadratic formula??
I'm sorry for my typo, I have edited it.
I have now edited in my steps, per request of commenters.

\begin{align}
(b-2)\sqrt{ac} & = \sqrt{(b-2)^2} \sqrt{ac} & & \text{ if } b-2\ge 0 \\[10pt]
& = \sqrt{(b-2)^2 ac}.
\end{align}
Is $(b-2)^2ac$ the same as $b^2-4ac\,$?
If one were to think that $(b-2)^2$ is the same as $b^2-4$ (and it is not) then one would have $(b-2)^2ac = b^2ac-4ac,$ so that is still not the same as $b^2-4ac.$
Notice that $\sqrt{5^2 - 3^2} = 4$ and $\sqrt{5^2}-\sqrt{3^2} = 5 - 3 = 2,$ so
$\sqrt{5^2-3^2}$ is different from $\sqrt{5^2}-\sqrt{3^2}.$
One should not generally ask why one cannot do things like this; but rather whether one can. Don't start from the presumption that it can be done. That puts the burden of proof in the wrong place.
