Answer
$t^2+\dfrac{17}{6}t+2$
Work Step by Step
Using $(a+b)(c+d)=ac+ad+bc+bd$ or the product of $2$ binomials, the given expression, $
\left( t+\dfrac{3}{2} \right)\left( t+\dfrac{4}{3} \right)
$, is equivalent to
\begin{array}{l}\require{cancel}
t(t)+t\left( \dfrac{4}{3} \right)+\dfrac{3}{2}(t)+\dfrac{3}{2}\left( \dfrac{4}{3} \right)
\\\\=
t^2+\dfrac{4}{3}t+\dfrac{3}{2}t+\dfrac{12}{6}
\\\\=
t^2+\dfrac{8}{6}t+\dfrac{9}{6}t+2
\\\\=
t^2+\dfrac{17}{6}t+2
.\end{array}
