Each solution corresponds to a factor of the equation, so the equation must look like this:
\begin{equation*}
\left(x-\dfrac{1}{2}\right)(x-(-3)) = 0
\end{equation*}
or, simplifying:
\begin{equation*}
\left(x-\dfrac{1}{2}\right)(x+3) = 0
\end{equation*}
We multiply the factors together to obtain
\begin{equation*}
x^2+\dfrac{5}{2}x-\dfrac{3}{2} = 0
\end{equation*}
This is an equation that works, but we can make a "nicer" one if we clear the fractions. We can multiply both sides of the equation by 2. We know that multiplying by a constant does not change the solutions of the equation.
\begin{align*}
\alert{2}\left(x^2+\dfrac{5}{2}x-\dfrac{3}{2}\right) \amp = \alert{2}(0)\\
2x^2+5x-3 \amp = 0
\end{align*}
By factoring, we can check that this equation really does have the given solutions.
\begin{equation*}
0 = 2x^2+5x-3 = (2x-1)(x+3)
\end{equation*}
From here, you can see that the solutions are indeed \(\dfrac{1}{2}\) and \(-3\text{.}\)