What basis and coordinate system is used in this quadratic Bézier triangle equation? $[x,y,z] = A*s^2 + B*t^2 + C*u^2 + D*2st + E*2tu + F*2su$ I have the following equation for a quadratic Bézier triangle, but I'm having a lot of trouble understanding how to describe it:
$[x,y,z] = A*s^2 + B*t^2 + C*u^2 + D*2st + E*2tu + F*2su$

Specifically, what is the basis and coordinate system? My ultimate goal is to convert it to implicit form, and I've spent the past few weeks reading research papers on the topic, but it seems I am still stuck on the absolute basics.
Is it a trivariate symmetric parametric equation in Bernstein form, with two independent variables in a barycentric coordinate system? If so, what does the power form or Cartesian coordinate system version look like? If not, how do I get it to barycentric Bernstein form?

The $A,B,C,D,E,F$ are the 6 control points. They are completely analogous to the 4 control points used to define a cubic Bézier curve. 
The $s,t,u$ things are barycentric coordinates within a standard domain triangle. They have the property that $s+t+u=1$.
The functions $s^2, t^2, u^2, 2st, 2tu, 2su$ are the basis functions. They are generalized Bernstein polynomials. The usual Bernstein polynomials, as used in quadratic Bézier curves, are $(1-t)^2$, $2t(1-t)$, $t^2$. But, if you write $s+t=1$, then these can be written as $s^2$, $2st$, $t^2$. This should help you see the connection.
Everything here is coordinate-free. So, if you use some rectangular coordinate system to express $A,B,C,D,E,F$, then the thing you're calling $[x,y,z]$ will be the coordinates of the point $P(s,t,u)$ on the surface, expressed in this same ( rectangular) coordinate system.
Implicitizing can be done. You eliminate $s,t,u$, typically by using resultants of one sort or another. If I remember correctly, you'll get an implicit equation of degree 4, which (I think) is a Steiner surface. There is paper by Sederberg and Anderson that discusses this. They implicitize to try to make ray-tracing easier/faster.
