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: 
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?
When they get to point of having
$$\bigl{(}x+\frac{b}{2a}\bigr{)}^2 =-\frac{c}{a}+\bigl{(}\frac{b}{2a}\bigr{)}$$ the square root of both sides has no absolute value sign on the left because the right side can be $\pm$.
$$\biggl{|}x+\frac{b}{2a}\biggr{|}\ne\sqrt{-\frac{c}{a}+\biggl{(}\frac{b}{2a}\biggr{)}}$$

One complete derivation explained simply is shown [here](https://www.mathsisfun.com/algebra/quadratic-equation-derivation.html).