Answer
$a^2+\dfrac{21}{10}a-1$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the product of $2$ binomials, the given expression, $
\left( a-\dfrac{2}{5} \right)\left( a+\dfrac{5}{2} \right)
$, is equivalent to
\begin{array}{l}\require{cancel}
a(a)+a\left( \dfrac{5}{2} \right)-\dfrac{2}{5}(a)-\dfrac{2}{5}\left( \dfrac{5}{2} \right)
\\\\=
a^2+\dfrac{5}{2}a-\dfrac{2}{5}a-\dfrac{10}{10}
\\\\=
a^2+\dfrac{25}{10}a-\dfrac{4}{10}a-1
\\\\=
a^2+\dfrac{21}{10}a-1
.\end{array}