n^2+n-400≤0 怎么解

n^2+n-400≤0 怎么解

1^2-4*1*(-400)=1601 有点难啊

n1=(-1+√1601)/2,n2=(-1-√1601)/2,

n^2+n-400≤0 =======> n2<n<n1
=======> (-1-√1601)=<2 =<n<(-1+√1601)/2,

配方：(n+1/2)^2-1/4-400<=0

n^2+n-400≤0

-1601/4 + (1/2 + n)^2≤0
1/2 (-1 - √[1601]) <= n <= 1/2 (-1 + √[1601])
-20.5062 <= n <= 19.5062
整数解:
-20 <= n <= 19