Answer
$30$ rows
Work Step by Step
The number of seats forms an arithmetic sequence, with
$a_{1}=10,d=4,S_{n}=2040$
The sum $S_{n}$ of the first $n$ terms of $\left\{a_{n}\right\}$:
$S_{n}=\displaystyle \frac{n}{2}\left[2a_{1}+(n-1)d\right]$ ... which we solve for n
$2040=\displaystyle \frac{n}{2}[2(10)+(n-1)4]$
$4080=n[20+4n-4]$
$4080=4n^{2}+16n$
$4n^{2}+16n-4080=0$
$n^{2}+4n-1020=0$
...use the quadratic formula
$n=\displaystyle \frac{-4\pm\sqrt{16-4(1)(-1020)}}{2(1)}$
... discard the negative solution
$n=\displaystyle \frac{-4+64}{2}=30$
