Answer
$q^2-\dfrac{3}{2}qr+\dfrac{9}{16}r^2$
Work Step by Step
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $
\left( q-\dfrac{3}{4}r \right)^2
,$ is equivalent to
\begin{array}{l}\require{cancel}
(q)^2+2(q)\left(-\dfrac{3}{4}r\right)+\left(-\dfrac{3}{4}r\right)^2
\\\\=
q^2-\dfrac{3}{2}qr+\dfrac{9}{16}r^2
.\end{array}