Square of $(7-4x)$ is:

Using the algebraic identity $(a-b{)}^2=a^2-2ab+b^2$, let $a=7$ and $b=4x$:
$$(7-4x{)}^2=(7{)}^2-2(7)(4x)+(4x{)}^2$$
$$(7-4x{)}^2=49-56x+16x^2$$
Rearranging the terms in descending order of power gives:
$$16x^2-56x+49$$