Given the inequality:
$$ -1.2^n- \frac{n}{5} < \frac{N}{20} -\frac{6}{5}$$
The value of $n$ doesn't have a closed form, but i'm interested in approximations or tips on how to solve it iteratively.
$N$ is any positive integer value, maybe can be 30 for this case and $n$ can be any positive value, not necessary be integer.
To generalize your question somewhat, suppose that we are looking for solutions to the equation $$f(n)=-a^n-bn<c,$$ where $a>1,$ and $b>0.$ I will also assume that you are looking for solutions that are real numbers. To put this in terms of the problem you are trying to solve specifically, just let $a=1.2$ and $b=.2$.
If the coefficients are in the range given above, then your function $f(n)$ is strictly decreasing. That tells us that if we can find some value $x$ where $f(x)=c$ then if $n<x$ we have that $f(n)>c$ and if $n>x$ then we have that $f(n)<c.$
Ok, so we just need to find the value $x$ where $f(x)=c$. There is a nice algorithm for solving equations like these known as Newton's method. You can go to https://en.wikipedia.org/wiki/Newton%27s_method for an explanation of how it works. Anyway, here is the algorithm. Start of with some guess $x_0$ which you think is reasonably close to solving the equation. So $f(x_0)\approx c$. It doesn't need to be a very good guess. Then let $x_1=x_0-\frac{f(x_0)-c}{f'(x_0)}.$ Then do it again, letting $x_2=x_1-\frac{f(x_1)-c}{f'(x_1)}.$ Continue this process until you have a decent approximation to a solution. It should converge pretty quickly.
I defined $f$ above, but I should also say that the derivative of $f$ is given by $f'(n)=-\ln(a)a^n-b.$ I know I haven't answered your question fully, but this should put you on the right track to figure it out. Have fun!