Answer
$t^2-\dfrac{2}{5}t+\dfrac{1}{25}$
Work Step by Step
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the given expression, $
\left( t-\dfrac{1}{5} \right)^2
$, is equivalent to
\begin{array}{l}\require{cancel}
(t)^2+2(t)\left( -\dfrac{1}{5} \right)+\left( -\dfrac{1}{5} \right)^2
\\\\=
t^2-\dfrac{2}{5}t+\dfrac{1}{25}
.\end{array}