Absolute value inequality

$\begin{aligned} \left| \frac{20}{n+2} \right| & > 5 \\ \implies \frac{400}{(n+2)^2} & > 25 \\ (n+2)^2 & < 16 \\ n^2 + 4n - 12 & < 0 \\ (n+6)(n-2) & < 0 \end{aligned}$

Since the inequality is undefined when $n = -2$ the acceptable integers $n$ are $-5$ , $-4$ , $-3$ , $-1$ , $0$ and $1$ . And their sum $-5-4-3-1+0+1 = \boxed{-12}$ .

Squaring both sides we have $\dfrac{400}{(n+2)^2}>25$ or $16>(n+2)^2$ for $n \neq -2$ .

So taking square root on both sides, we have $|n+2|<4$ for $n \neq -2$ .

If $n$ is non-negative, we have $n+2<4$ or $n<2$ . In this case $n$ can be $0$ or $1$ .

If $n$ is negative, we have $-n-2<4$ or $n<-6$ for $n \neq -2$ . In this case $n$ can be $-5,-4,-3$ or $-1$ .

Therefore sum of all integer values of $n$ is $-5-4-3-1+0+1=\boxed{-12}$ .

Relevant wiki: Absolute Value Inequalities - Problem Solving

We can make use of the property: $\left \vert \frac{x}{y} \right \vert = \frac{|x|}{|y|}$ if $y \neq 0$ . Note that $n$ cannot be $-2$ since that that would make the denominator equal to zero.

$\phantom{\implies} \dfrac{|20|}{\lvert n + 2 \rvert } > 5$

We can divide both side of the inequality by the positive number 5.

$\implies \dfrac{4}{\lvert n + 2 \rvert } > 1$

The fraction is greater than one, if and only if the numerator is greater than the denominator.

$\phantom{\implies} \quad |n + 2| < 4$

$\implies -4 < n+2 < 4$

$\implies -6 < n < 2$

The integers satisfying this inequality are $\{-5, -4, -3, -1, 0, 1 \}$ . The sum of these values is $\boxed{-12}$ $_\square$