Answer
$\dfrac{1}{4}x^2 -9$
Work Step by Step
Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, then,
\begin{array}{l}
\left( \dfrac{1}{2}x+3 \right)\left( \dfrac{1}{2}x-3 \right)
\\\\=
\left( \dfrac{1}{2}x\right)^2 - (3)^2
\\\\=
\dfrac{1}{4}x^2 -9
.\end{array}
