An algebra problem by Ankit Nigam

$\displaystyle \sum^{n}_{k=1}(x+k-1)(x+k)=10n \\ \displaystyle \sum^{n}_{k=1}(x^2+2kx-x+k^2-k)=10n \\ nx^2+2\cdot\frac{n(n+1)}{2}x-nx+\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}-10n=0 \\ \text{We will finally get } 3x^2+3nx+n^2-31=0 \\ \text{By Vieta's: } 2\alpha+1=-n \Rightarrow \therefore \alpha=-\frac{(n+1)}{2} \\ \alpha(\alpha+1)=\frac{n^2-31}{3}\Rightarrow \alpha^2+\alpha=\frac{n^2-31}{3} \\ \frac{(n+1)^2}{4}-\frac{(n+1)}{2}=\frac{n^2-31}{3} \\ 3n^2+3+6n-6n-6=4n^2-124 \\ n=\boxed{11}$