## Algebra 1

$n = 5$
$280 = 4*(n+2)*(n+5)$ $280/4 = 4*(n+2)*(n+5)/4$ $70 = (n+2)(n+5)$ $n^2+2n+5n+10 = 70$ $n^2+7n+10-70 = 70-70$ $n^2+7n-60 = 0$ $(n+12)(n-5) = 0$ $n+12 = 0$ $n+12-12 = 0-12$ $n = -12$ $n-5 = 0$ $n-5+5 = 0+5$ $n = 5$ We can't have a negative length, so $n\ne-12$.