Answer
$\dfrac{3}{16}a^2+\dfrac{5}{4}a-2$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the product of $2$ binomials, the given expression, $
\left( \dfrac{1}{4}a+2 \right)\left( \dfrac{3}{4}a-1 \right)
$, is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{4}a\left( \dfrac{3}{4}a \right)+\dfrac{1}{4}a(-1)+2\left( \dfrac{3}{4}a \right)+2(-1)
\\\\=
\dfrac{3}{16}a^2-\dfrac{1}{4}a+\dfrac{6}{4}a-2
\\\\=
\dfrac{3}{16}a^2+\dfrac{5}{4}a-2
.\end{array}
