Question 250634
Hi, lets see if we can get you going in the right direction.

If you have {{{sqrt(110-n)}}} = n and square both sides you end up with

110-n = {{{n^2}}}.

(Hopefully I am editing this so it appears right) 
It sounds like you are stuck at this point but you are on the right track. If you put everything on the right side with {{{n^2}}} you will have a quadratic equation.  There are several ways to solve this but the easiest way to solve this quadratic equation is by factoring. Here are the steps:

Add n to both sides, subtract 110 from each side to get
0 = {{{n^2}}} + n - 110
Now you can solve this by factoring  {{{n^2}}} + n - 110.
You need to find factors of 110 that add to be 1. These factors are -10 and +11.
Factoring the trinomial gives you the equation 0 = (n - 10)(n + 11).
Solving this involves setting each factor equal to zero and solving.
n - 10 = 0           n + 11 = 0
n = 10 and -11.
Plug the solutions back into the equation to find out if they are solutions:
{{{sqrt(110-10)}}}=10  |  {{{sqrt(100)}}}=10 True
{{{sqrt(110+11)}}}=11  |  {{{sqrt(121)}}}=11 True

n = 10 and -11 are the solutions.

I hope this gets you working in the right direction.