For example, if {$a_0, a_1, a_2, a_3, ...$} is the sequence, the first difference is {$a_1-a_0, a_2-a_1, a_3-a_2, ...$}, and the second difference is {$(a_2-a_1)-(a_1-a_0), (a_3-a_2)-(a_2-a_1), ...$}.
I think that using facts from up to Calculus, perhaps derivatives, should be enough. I find myself going in circles and don't know how to approach this.
Let $d$ denote the constant second difference. Moreover, let \begin{align*} c\equiv&\,a_1-a_0-\frac{d}{2},\\ \end{align*}
Proof: The claim is obviously true for $n=0$. For $n=1$, $$\frac{d}{2}\times n^2+cn+a_0=\frac{d}{2}+\left(a_1-a_0-\frac{d}{2}\right)+a_0=a_1.$$ Proceed by induction: suppose that the claim is true for $0,1,\ldots,n$ for some integer $n\geq1$. The task is to prove that it is true for $n+1$. Now: \begin{align*} d=&\,(a_{n+1}-a_n)-(a_n-a_{n-1})=a_{n+1}-2a_n+a_{n-1}\\ =&\,a_{n+1}-2\left[\frac{d}{2}\times n^2+cn+a_0\right]+\left[\frac{d}{2}\times (n-1)^2+c(n-1)+a_0\right], \end{align*} where the first equality comes from the definition of $d$, and the third one is due to the induction hypothesis. Now, rearrange for $a_{n+1}$: \begin{align*} a_{n+1}=&\,d+2\left[\frac{d}{2}\times n^2+cn+a_0\right]-\left[\frac{d}{2}\times (n-1)^2+c(n-1)+a_0\right]\\ =&\,d+dn^2+2cn+2a_0-\frac{d}{2}\times(n^2-2n+1)-c(n-1)-a_0\\ =&\,\underbrace{d}_{\spadesuit}+\underbrace{dn^2}_{\heartsuit}+\underbrace{2cn}_{\clubsuit}+\underbrace{2a_0}_{\diamondsuit}-\underbrace{\frac{d}{2}\times n^2}_{\heartsuit}+\underbrace{dn}_{\star}-\underbrace{\frac{d}{2}}_{\spadesuit}-\underbrace{cn}_{\clubsuit}+\underbrace{c}_{\clubsuit}-\underbrace{a_0}_{\diamondsuit}\\ =&\,\underbrace{\frac{d}{2}}_{\spadesuit}+\underbrace{\frac{d}{2}\times n^2}_{\heartsuit}+\underbrace{c(n+1)}_{\clubsuit}+\underbrace{a_0}_{\diamondsuit}+\underbrace{dn}_{\star}\\ =&\,\frac{d}{2}\times n^2+dn+\frac{d}{2}+c(n+1)+a_0\\ =&\,\frac{d}{2}\times(n+1)^2+c(n+1)+a_0. \end{align*} The proof is complete. $\quad\blacksquare$