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?

That's a good question. 
But for the sake of the quadratic equation we don't need $\sqrt{x/y^2} = \sqrt{x}/y$.  We just need $\sqrt{x/y^2} = \pm \sqrt{x}/y$. 
If $y^2 = c \ne 0$ then $y = \{x_1,x_2\}$ where $x_1 = -x_2$.  This is true for all $c$, real or complex.  Arbitrarily declare one of them, say $x_1$, to be the $\sqrt{c}$ then we can always have $\sqrt{y^2} = \pm y$.  Always. 
(If $y^2 = 0 \implies y = 0$ and $0 = \pm 0$.)
If $y^2 = c = r*e^{i\theta} \in \mathbb C$ then $y = \{\sqrt{r}e^{i(\theta/2},\sqrt{r}e^{i\theta/2 + \pi}\}$ = $\pm \sqrt{r} e^{i\theta/2}$
So $\sqrt{x/y^2} = \pm \sqrt{x}/y$.  So equation is good even for complex numbers.
