Find the roots of the equation - $z^2 +12jz+64 = 0$

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Just needing a little guidance. This is what I've done so far and I'm not sure if I'm doing it right.

Using quadratic formula:

$$z^2+12jz+64=0$$

$$ z= \frac{-12j ±\sqrt{(12j^2-4\times1\times64)}}{2\times1}=\\ -12j ±\frac{\sqrt{3072}}{2}=\\ -12j ±\frac{2\sqrt{1536}}{2}=\\ -12j±\sqrt{1536}.$$

Is this right way to go about it? Thanks in advance.

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Assuming $\;j=i=\sqrt{-1}\;$ :

$$z^2+12iz+64=0\implies \Delta=b^2-4ac=-144-256=-400=(20i)^2\implies$$

$$z_{1,2}=\frac{-12i\pm20i}{2}=\begin{cases}-16i\\{}\\\;\;\;\,4i\end{cases}$$