How do I find complex roots of a quartic polynomial using quadratic formula? For something like:
$$
z^4 + 8z^2 + 3
$$
how can I find all the complex roots using the quadratic equation, or is there a better method?
I tried a u substitution using $u = z^2$, but then when I applied the quadratic formula, I was getting real roots and there was no where to sub u back in.

Presumably you got $u=z^2=-4\pm\sqrt{13}$.  Now you need to solve for $z$.
Both of our values of $z^2$ are negative. The solutions of $z^2=-4+\sqrt{13}$ are 
$z=\pm i \sqrt{4-\sqrt{13}}$, and the solutions of $z^2=-4-\sqrt{13}$ are $z=\pm i\sqrt{4+\sqrt{13}}$. 
Remark: Things get somewhat more complicated when you want to find the square roots of a general complex number. For example, suppose that we want to find the square roots of $3+4i$. One usual way is to first rewrite $z$ as
$$\sqrt{3^2+4^2} \left( \frac{3}{\sqrt{3^2+4^2}}+ \frac{4}{\sqrt{3^2+4^2}}i\right).$$
Note that $\sqrt{3^2+4^2}=5$.
Let $\theta$ be an angle whose cosine is $\frac{3}{5}$ and whose sine is $\frac{4}{5}$. Then the square roots of $z$ are
$$\pm \sqrt{5}\left(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}i\right).$$ 
