Universal sums of three quadratic polynomials

Let $a,b,c,d,e$ and $f$ be integers with $a\ge c\ge e>0$, $b>-a$ and $b\equiv a\pmod2$, $d>-c$ and $d\equiv c\pmod 2$, $f>-e$ and $f\equiv e\pmod2$. Suppose that $b\ge d$ if $a=c$, and $d\ge f$ if $c=e$. When $b(a-b)$, $d(c-d)$ and $f(e-f)$ are not all zero, we prove that if each $n\in\mathbb N=\{0,1,2,\ldots\}$ can be written $x(ax+b)/2+y(cy+d)/2+z(ez+f)/2$ with $x,y,z\in\mathbb N$ then the tuple $(a,b,c,d,e,f)$ must be on our list of $473$ candidates, and show that 56 of them meet our purpose. When $b\in[0,a)$, $d\in[0,c)$ and $f\in[0,e)$, we investigate the universal tuples $(a,b,c,d,e,f)$ over $\mathbb Z$ for which any $n\in\mathbb N$ can be written $x(ax+b)/2+y(cy+d)/2+z(ez+f)/2$ with $x,y,z\in\mathbb Z$, and show that there are totally 12082 such candidates some of which are proved to be universal tuples over $\mathbb Z$. For example, we show that any $n\in\mathbb N$ can be written as $x(x+1)/2+y(3y+1)/2+z(5z+1)/2$ with $x,y,z\in\mathbb Z$, and conjecture that each $n\in\mathbb N$ can be written as $x(x+1)/2+y(3y+1)/2+z(5z+1)/2$ with $x,y,z\in\mathbb N$.