Step 1: Use the functional equation at a special point.
Put $x=y=0$ in $f(x+y)=f(x)+f(y)+xy$ to get $f(0)=2f(0)$, so $f(0)=0$. Since $f(0)=c$, this means $c=0$.
Step 2: Compare the $xy$ term.
Expanding $f(x+y)=a(x+y)^2+b(x+y)$ gives a cross term $2axy$; matching the $xy$ on the right forces $2a=1$, so $a=\dfrac12$.
Step 3: Use the side condition.
Given $a+b+c=3$ with $a=\dfrac12$, $c=0$, we get $b=3-\dfrac12=\dfrac52$.
Step 4: Write the explicit function.
Thus $f(n)=\dfrac12 n^2+\dfrac52 n$.
Step 5: Sum from $1$ to $10$.
$\displaystyle\sum_{n=1}^{10} f(n)=\dfrac12\sum n^2+\dfrac52\sum n$, where $\sum_{1}^{10} n^2=385$ and $\sum_{1}^{10} n=55$.
Step 6: Evaluate.
This is $\dfrac12(385)+\dfrac52(55)=\dfrac{385}{2}+\dfrac{275}{2}=\dfrac{660}{2}=330$.
\[ \boxed{330} \]