10(x+1)/(x^2+2x+3) >= 1
since x^2+2x+3 > 0,
10x+10 >= x^2+2x+3
x^2 - 8x - 7 <= 0
So, all integers n such that
4-√23 <= n <= 4+√23
How many integers satisfy the inequality
\left| \frac { 10(x+1) } {x^2 + 2x + 3 } \right| \geq 1?
1 answer