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.
Because $\sqrt{b^2 - 4ac}\neq 2b\sqrt{ac}$. Say for example that $c = 0$ and $b\neq 0$. Then you have
\begin{align*}
\sqrt{b^2 - 4ac} &amp;= \sqrt{b^2}\\
&amp;= \left|b\right|\\
&amp;\neq 0,
\end{align*}
so you can see that this simplification cannot be correct.

It seems that in your proposed simplification, you have completely disregarded the subtraction occurring in the radical. Moreover, $\sqrt{b^2} = \left|b\right|$, not just $b$. To see this with an example, take $b = -1$. Then $\sqrt{(-1)^2} = \sqrt{1} = 1 = \left|-1\right|\neq -1$.



EDIT:

Again, the simplification is incorrect. While it is true in general that $\sqrt{a^2b} = \left|a\right|\sqrt{b}$, this is not the situation you are in here:
$$
b^2 - 4ac\neq (b - 2)^2 ac = (b^2 - 4b + 4)ac = b^2 ac - 4abc + 4ac.
$$
You seem to have made a few mistakes here (if I'm to take a stab at the reasoning behind the simplification): first you've incorrectly simplified $b^2 - 4ac$ as $(b^2 - 4)ac$ (which is not true, because the first term in the former has no $ac$), and then you've simplified $b^2 - 4$ as $(b - 2)^2$, which is also not true (take $b = 0$ to see why). In general, $(x + y)^n\neq x^n + y^n$: this is a common mistake algebra learners make! Remember that when expanding $(x + y)^2$, we need to use the distributive property, and not simply regard squaring as linear:
\begin{align*}
(x + y)^2 &amp;= (x + y)(x + y)\\
&amp;= x^2 + yx + xy + y^2\\
&amp;= x^2 + 2xy + y^2.
\end{align*}
I think a better way to parse the quadratic equation is as:

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

The left term ($-\frac{b}{2a}$) is the x-coordinate of the axis of symmetry (often taught in Algebra as a formula, but easily recognized to students of calculus as the solution to the first-order condition of a quadratic).

The right-hand term is the "margin" around the axis of symmetry giving how far to the left/right of the axis we find the zeroes.

The numerator of the right-hand term is the discriminant, the sign of which determines the number of zeroes there are. 

The denominator of the right-hand term shows that as $a$ increases, ceteris paribus, the zeroes will move closer to the axis of symmetry -- intuitively, $a$ is the "stretch" factor, and higher $a$ means the parabola will be more "squeezed" (relatively speaking). The more "squeezed" the parabola, the closer its zeros will be to the "middle".