Answer
$t^2-\dfrac{9}{16}$
Work Step by Step
Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the given expression, $
\left( t-\dfrac{3}{4} \right)\left( t+\dfrac{3}{4} \right)
$, is equivalent to
\begin{array}{l}\require{cancel}
(t)^2-\left( \dfrac{3}{4} \right)^2
\\\\=
t^2-\dfrac{9}{16}
.\end{array}