Lorentz Transformation at t=0 Suppose I have two reference frames $S$ and $S'$, where $S'$ is moving with velocity $v$ with respect to $S$.
The Lorentz transformation equation for time in reference frame $S$ is given by:
$$t'=\gamma\left(t-x\frac{v}{c^2}\right)$$
where $\gamma$ is the Lorentz factor.
Now for an event happening at $t=0$ and at some position $x$ in $S$ frame, then $t'<0$ for $S'$ frame. But if $x=0$ then $t'=0$.
How should I interpret this result? 
 A: The interpretation is that two events being simultaneous as measured in frame $S$ doesn't imply that the events are simultaneous in frame $S'$.  Which events count as being "simultaneous" depends on the frame of reference.  This is known as the relativity of simultaneity.
Added clarification due to comment:
The coinciding of the origins is an event, call it event $A$.  The coordinates of $A$ as measured in $S$ are $t_A=0$, $x_A=0$, and the coordinates of $A$ as measured in $S'$ are $t'_A=0$, $x'_A=0$.
You're also considering another event, call it $B$, whose coordinates as measured in $S$ are $t_B=0$, $x_B=x$.  As measured in $S'$, the $t'$ coordinate of $B$ is $t'_{B}=-\gamma x v/c^2$.
$t'_B$ being negative (if $x>0$) means that according to $S'$, event $B$ occurs before event $A$.  However, if $x=0$, then $B$ is the same event as $A$, so $t'_B=0$ and $x'_B=0$.
There is no physical discontinuity between the situations where $x=0$ and $x>0$, i.e., between the situations where $t'_B=0$ and $t'_B<0$.  It's just that the further away event $B$ is from event $A$ spatially, the bigger the difference will be between $t'_B$ and $t_B$.
