How do you solve \[\left( 3n+2 \right)\left( n-2 \right)=0\]?

Hint: For the given question we have been asked to solve for n that is we have to find the value of n which satisfies the equation. To solve the questions of this kind we will equate each term inside the bracket to zero. After doing so we will simplify the equations further to get the solution to the given question.

Complete step by step answer:
Firstly, for the equation \[\left( 3n+2 \right)\left( n-2 \right)=0\] we will equate the two terms to zero.
So, in the first case we will equate the term in the first bracket to zero. So, we get the equation reduced as follows.
\[\Rightarrow \left( 3n+2 \right)=0\]
Here we will send the \[2\] to the right hand side of the equation. So the equation will be reduced as follows.
\[\Rightarrow 3n=-2\]
Now, we will send the \[3\] to the other side of the equation so that it becomes the denominator on the other side. So, we get the equation reduced as follows.
\[\Rightarrow n=-\dfrac{2}{3}\]
Now, for case two we will equate the second bracket term in the equation \[\left( 3n+2 \right)\left( n-2 \right)=0\] to zero.
So, we get the equation as follows.
\[\Rightarrow n-2=0\]
Now, we will send integer \[-2\] to the other side that is the right hand side of the equation.
\[\Rightarrow n=2\]
So, the resultant value of n we got in both cases is our solution.

Therefore, the solution will be \[n=2,-\dfrac{2}{3}\].

Note: Students must be very careful in doing the calculations. We should know that to find the solutions to these questions we have to equate the terms to zero. Here we should not do calculation mistakes like for example in case one if we write the step
as \[ 3n=2\] instead of \[ 3n=-2\] our solution will be a wrong one.