Answer
$\dfrac{6}{25}t^2-\dfrac{1}{5}t-1$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the product of $2$ binomials, the given expression, $
\left( \dfrac{2}{5}t-1 \right)\left( \dfrac{3}{5}t+1 \right)
$, is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{2}{5}t\left( \dfrac{3}{5}t \right)+\dfrac{2}{5}t(1)-1\left( \dfrac{3}{5}t \right)-1(1)
\\\\=
\dfrac{6}{25}t^2+\dfrac{2}{5}t-\dfrac{3}{5}t-1
\\\\=
\dfrac{6}{25}t^2-\dfrac{1}{5}t-1
.\end{array}
