Question: When deriving the quadratic formula, isn't the square root of $(x+\frac{b}{2a})^2$ the absolute value of $(x+\frac{b}{2a})$? It's usually just represented as $(x+\frac{b}{2a})$ without absolute value and then $\frac{b}{2a}$ is subtracted from left side and boom, theres the quadratic formula. I just don't understand why its not absolute value of $(x+\frac{b}{2a})$. For example the square root of $x^2$ is the absolute value of $x$, which is equal to $\pm{x}$. Sorry if this is confusing its more of a conceptual thing. Thank you in advance.

philalethesnew

Answer: 
You seem to be asking why after the step $\;\left(x+\cfrac b{2a}\right)^2=\cfrac{b^2-4ac}{4a^2}\;$ , they don't go to

$$\left|x+\cfrac b{2a}\right|=\sqrt{\frac{b^2-4ac}{4a^2}}$$

But we **actually** do! To write the above is **exactly** the same as to write

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

just as $\;x^2=a\implies |x|=a\;\; (\,a\ge 0)\;$ is exactly the same as $\;x^2=a\implies x=\pm a\;\;(a\ge0)\;$ , under the assumption, of course, that once we take the + sign and the second time we take the - sign.
Well actually it is done. 

$$ax^2 + bx + c = 0$$

$$\implies x^2+\frac{bx}{a} +\frac{c}{a} = 0$$

Now completing the square: 

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

Now since we know that, 

$$x^2 = a, \quad a > 0$$ 

has two solutions, $x=\pm a$. 

Now if $b^2 - 4ac > 0$, there are two solutions indeed! 

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

Does that help?