If I have a tank of water filled to depth H and if I let $z$ measure the distance from the bottom of the tank at $z=0$.
The fluid has density $ρ$ and the tank has a small hole of area $a$ in the base of which water jets out of.
If we now assume that the rate at which the level of water in the tank decreases is negligible in comparison to the speed of the jet.
I am looking to find the speed in the jet.
I am trying to use bernoulli's equation to solve this however I don't quite understand how to.
$\textit{Bernoulli}$ holds along any streamline.
Let take two points (that link two points of the streamline in question), one on an arbitrary point of the free surface of the liquid, and another one in the hole. Hence
$$\frac{p_s}{\rho} + gH + \tfrac{1}{2}v_s^2 = \frac{p_h}{\rho} + \tfrac{1}{2}v_h^2$$
Being the subscripts $s$ and $h$ belong to the free surface and the hole respectively.
It is clear that the velocity on the surface is almost 0 (related with your assumption of the velocity of the jet), hence $v_s\approx 0$, and iff the tank is open, then $p_s=p_h$.
Apply this reasoning to obtain $v_h$