How do you find $p\left( t-2 \right)$ given $p\left( t \right)=4t-5$?

Hint: We have been a function $p$ in t-variable. This function is a linear equation in one variable, t which consists of a term in t-variable and a constant term. We have to find the value of $p\left( t-2 \right)$which we will calculate by substituting $t-2$ in place of $t$ in the given function. Further, we shall open the brackets and multiply each term by 4 to perform the required addition or subtraction of constant terms to finally simplify the equation.

Complete step by step solution:
Given that $p\left( t \right)=4t-5$.
This is a linear equation in one-variable, t-variable and which has a term 4t in t-variable and a constant term only.
Now, the expression $p\left( t-2 \right)$ represents the same function$p$in variable-$\left( t-2 \right)$ instead of the variable-t. Another way of understanding it would be that the given function is simply written in terms of $\left( t-2 \right)$ instead of variable-t.
Thus, in order to find $p\left( t-2 \right)$, we shall substitute variable-t with $\left( t-2 \right)$ at all places where variable-t is written in the function.
$\Rightarrow p\left( t \right)=4t-5$
Substituting variable-t with $\left( t-2 \right)$, we get
$\Rightarrow p\left( t-2 \right)=4\left( t-2 \right)-5$
We shall now open the brackets of $\left( t-2 \right)$ and multiply each term by 4.
$\Rightarrow p\left( t-2 \right)=4t-8-5$
Now, we will add both the negative constant terms to simplify the expression more.
$\Rightarrow p\left( t-2 \right)=4t-13$
Therefore, if $p\left( t \right)=4t-5$, then $p\left( t-2 \right)$ is equal to $4t-13$.

Note:
One possible mistake we could have done while solving this problem would be while opening the brackets of the formed new expression. Sometimes, we tend to ignore the negative sign used between the terms within such brackets and write them as positive terms. However, doing so would definitely lead us to incorrect results.