When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$? The solution to the quadratics is given by
$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to
$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and $\sqrt{\dfrac{x}{y}} \neq \dfrac{\sqrt{x}}{\sqrt{y}}$ if $x$ and $y$ are negative, but given that we have $a^2$, is this justified?
Is $\sqrt{x/y^2} = \sqrt{x}/y$? Is $r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ always true?

$$\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$$
Is "the solution" to $ax^2 + bx + c$, provided that $a,b,c$ are real numbers and $a \ne 0$.
$4a^2 \ge 0$ for all real numbers $a$. Since we are excluding $a = 0$, then $4a^2 > 0$. We can then say 
$$\sqrt{4a^2} = 
\begin{cases}
    2a,  & \text{if $a \ge 0$} \\
    -2a, & \text{if $a < 0$}
\end{cases}$$
The real problem has to do with $b^2 - 4ac$ since it can be positive, negative, or $0$. But, in all three cases, we will have
$$\sqrt{\dfrac{b^2-4ac}{4a^2}}= \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
so we end up with
\begin{align}
    \dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}
    &= \dfrac{b}{2a}\pm \left( \pm \dfrac{\sqrt{b^2-4ac}}{2a} \right) \\
    &= \dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a} \\
\end{align}
By the way, for any real number $x$, $\sqrt{x^2} = |x|$.
