How do you solve $ \left| -2n+6 \right|=6$?

Hint: We explain the term absolute value of a number. How the modulus function always remains positive. We expand the function and break it into two parts. Then we verify the result with an example.

Complete step by step answer:
Modulus function $ f\left( x \right)=\left| x \right|$ works as the distance of the number from 0. The number can be both positive and negative but the distance of that number will always be positive. The distance can never be negative.
In mathematical notation we express it with modulus value. Let a number be x whose sign is not mentioned. The absolute value of that number will be $ \left| x \right|$. We can say $ \left| x \right|\ge 0$.
We can express the function $ f\left( x \right)=\left| x \right|$ as $ f\left( x \right)=\left\{ \begin{align}
  & x\left( x\ge 0 \right) \\
 & -x\left( x<0 \right) \\
\end{align} \right.$.
We can write $ f\left( x \right)=\left| x \right|=\pm x$ depending on the value of the number x.
For our given function we break the value of n into two parts. One being $ n>3$ and other being $ n\le 3$.
In case of $ n>3$, the function becomes \[\left| -2n+6 \right|=-\left( -2n+6 \right)=2n-6\].
In case of $ n\le 3$, the function becomes \[\left| -2n+6 \right|=6-2n\].
The solution for $ n>3$, \[2n-6=6\Rightarrow n=\dfrac{6+6}{2}=6\].
The solution for $ n\le 3$, \[6-2n=6\Rightarrow n=0\].
Therefore, the solutions are $ n=0,6$.

Note:
The only time the absolute value becomes 0 is when the number itself is 0. For any other number the absolute value is greater than 0. Therefore, we can say $ \left| x \right|>0$ when $ x\ne 0$. We also can solve as
$\begin{align}
  & \left| -2n+6 \right|=6 \\
 & \Rightarrow -2n+6=\pm 6 \\
 & \Rightarrow -2n=\pm 6-6=0,-12 \\
 & \Rightarrow n=0,6 \\
\end{align}$