Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 103: 115

$37$

$\displaystyle \frac{1}{2}n(n+1)=703\quad/\times 2$ $n^{2}+n=1406$ $n^{2}+n-1406=0$ $a=1,b=-1,c--1406$ $n=\displaystyle \frac{-1\pm\sqrt{1+4(1)(1406)}}{2} =\frac{-1\pm 75}{2}$ Discarding the negative, $n=\displaystyle \frac{-1+75}{2}=37$