I've a quadratic equation:
$$f(m)=am^2+bm+c\tag1$$
Where $a,b,c\in\mathbb{R}$ (where $c$ can be zero), $m\in\mathbb{N}$ and $m\ge2$.
Is there a clever way of solving (where $\text{n}\in\mathbb{N}$ and $\text{n}\ge2$):
$$\sum_{k=1}^\text{n}f(k)=f(m)\space\Longleftrightarrow\space m=\dots\tag2$$
As an example we can look at: $f(m)=-\dfrac{m}{2}+\dfrac{3m^2}{2}$
You are asking to solve the Diophantine equation
$$a\frac{n(n+1)(2n+1)}{6}+ b\frac{n(n+1)}{2}+ cn=am^2+bm+c.$$
If some of the coefficients $a,b,c$ are incommensurable, the equation turns to a system. E.g.
$$\sqrt2\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\sqrt2m^2+m$$ is equivalent to
$$\begin{cases}\dfrac{n(n+1)(2n+1)}{6}=m^2,\\\dfrac{n(n+1)}{2}=m\end{cases}$$
and a solution can exist if
$$\frac{n(n+1)(2n+1)}{6}=\dfrac{n^2(n+1)^2}{4}.$$
The integer solutions are $n=-1,0,1$.
When the coefficients are commensurable, you can turn them to integers and I guess that the resulting equation is difficult.