How do you simplify \[\left( 7-6i \right)\left( 2-3i \right)\]?

Hint: We first explain the meaning of the process ‘FOIL’. We multiply the terms according to their positions. There are four multiplications to be done. We complete all four multiplications according to the previously mentioned process.

Complete step by step solution:
We have been given multiplication of two linear equations. We have to do the breakings of the polynomials in order of FOIL. The word FOIL stands for First-Outside-Inside-Last. It is a technique to distribute the multiplication of polynomials.
There are two terms in each polynomial.
We start by multiplying the first terms of \[\left( 7-6i \right)\] and \[\left( 2-3i \right)\]. The terms are 7 and 2.
The multiplication gives a result of $7\times 2=14$.
We now multiply the outside terms of \[\left( 7-6i \right)\] and \[\left( 2-3i \right)\]. The terms are 7 and $-3i$.
The multiplication gives a result of $7\times \left( -3i \right)=-21i$.
Then we multiply the inside terms of \[\left( 7-6i \right)\] and \[\left( 2-3i \right)\]. The terms are $-6i$ and 2.
The multiplication gives the result of $\left( -6i \right)\times 2=-12i$.
We end by multiplying the last terms of \[\left( 7-6i \right)\] and \[\left( 2-3i \right)\]. The terms are $-6i$ and $-3i$.
The multiplication gives the result of $\left( -6i \right)\times \left( -3i \right)=18{{i}^{2}}$.
Now we add all the terms to get $\left( 7-6i \right)\left( 2-3i \right)=14-21i-12i+18{{i}^{2}}$.
We have the relations for imaginary $i$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. We place the values in the multiplication.
The final solution is $14-21i-12i+18{{i}^{2}}=14-33i-18=-33i-4$

Therefore, multiplied value of \[\left( 7-6i \right)\left( 2-3i \right)\] is $-33i-4$.

Note: We can find that in the multiplication the real numbers are created from the multiplication of two real or two imaginary numbers and the imaginary numbers are created from the multiplication of mixed numbers.