Keep it up

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

I did it using telescoping series . The given sum is a telescoping series!

Multiply And divide the summation by 3 . Then in numerator write 3 as (x+k+1) - (x+k-2) .

You will observe the telescoping series. which on simplification gives you the same quadratic equation which

Akshat Sharda Has got by apply sigma formula

Further use the fact that difference of roots of that quadratic is 1 and done! :)