Why is O2 a biradical?
We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for $\ce{O_2}$

Since an oxygen atom has 6 electrons,

A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, each oxygen would only have 7 electrons
B would correspond to a structure with a double bond between the oxygen atoms, 2 lone pairs on each oxygen and no unpaired electrons on each oxygen; B does have an octet around each oxygen, but it is not a biradical
C would correspond to a structure with a triple bond between the oxygen atoms, 1 lone pair on each oxygen and an unpaired electron on each oxygen; however C does not have an octet around each oxygen, in fact, each oxygen would have 9 electrons and this would be impossible for oxygen

So while structure A would indicate a biradical, we wouldn't &quot;expect&quot; it to count for much since the oxygens do not have octets.  This inability to clearly predict the biradical nature of $\ce{O_2}$ illustrates one of the failings of both Lewis structures and resonance theory.
In order to correctly predict the biradical nature of $\ce{O_2}$ we must move up to molecular orbital theory.  Below is the molecular orbital diagram for $\ce{O_2}$.  As you can see it does predict that $\ce{O_2}$ should be a biradical with an unpaired electron in each of its degenerate, highest occupied molecular orbitals.

Edit: response to OP's comment

When I think of triple bond I don't think of 2 3 electron bonds(which
is what you drew). Rather I think of 3 2 electron bonds(1 sigma bond
and 2 pi bonds)

Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.

This type of triple bond would make the oxygen positive with 5
electrons around it.

No, the formal charge on the oxygen in structure C is
Z = 6 - 3 unshared - (1/2 * 6 shared)= 0,
there is no formal charge on oxygen in the &quot;triple bond&quot; structure and as I noted above, there are 9 electrons around it (not 5), which is impossible for oxygen.

I am assuming electrons are shared equally with half around 1 atom and
half around the other(which is the basis for formal charge

Yes, that's correct.