How‌ ‌do‌ ‌you‌ ‌perform‌ ‌the‌ ‌operation‌ ‌and‌ ‌write‌ ‌the‌ ‌result‌ ‌in‌ ‌standard‌ ‌form‌ ‌given‌ ‌$\left(‌ ‌6-2i‌ ‌\right)\left(‌ ‌2-3i‌ ‌\right)$‌ ‌?‌

Hint: We are asked to simplify and write $\left( 6-2i \right)\left( 2-3i \right)$into standard form. To do so we have to understand what is the standard form of the complex number. We will learn what ‘i’ stands for. Once we get all these we will learn that while multiplying two complex numbers we just multiply each term of the first complex number by another complex number and then arrange them in standard form.

Complete step-by-step solution:
We are given two numbers as $6-2i$ and $2-3i$ we have to find the product of these.
We can say that these are complex numbers. The standard form of complex number is $a+ib$ denoted as $a+ib$ where ‘a’ and ‘b’ are real numbers while ‘i’ stand for iota denoted for $i=\sqrt{-1}$ .
When we multiply we multiply just as we do for real.
Now we also need to know how iota behaves when multiplied.
So, ${{i}^{2}}=-1,{{i}^{3}}=-i\text{ and }{{i}^{4}}=1$ and so on.
Now we start working on our problem.
We have $\left( 6-2i \right)\left( 2-3i \right)$
So, $\left( 6-2i \right)\left( 2-3i \right)=6\left( 2-3i \right)-2i\left( 2-3i \right)$
Opening brackets, we get –
$=6\times 2+6\times \left( -3i \right)+\left( -2i \right)\left( 2 \right)+\left( -2i \right)\left( -3i \right)$
Simplifying we get –
$=12+\left( -18i \right)-4i+6{{i}^{2}}$
Simplifying further we get –
$=12-18i-4i-6$ $\left[ \text{as }{{i}^{2}}=-1 \right]$
So,
$=6-22i$ $\left[ \text{as }12-6=6\text{ and }-18i-4i=22i \right]$
So we get –
$\left( 6-2i \right)\left( 2-3i \right)=6-22i$

Note: While simplifying, remember we can just add real with real and iota parts with iota part we can never add real with imaginary (iota part).
That is $2+3i+3i\ne 5i+3i$
We can add $2i+3i=5i$
We need to be careful around there. Also when we multiply we need to be careful with signs as it may contain many signs. So we need to understand that $\left( + \right)\left( - \right)=-\text{ and }\left( - \right)\left( - \right)=+\text{ and }\left( + \right)\left( + \right)=+$ .