Answer
$x^2-48=0$
Work Step by Step
We are given the solutions of a quadratic equation:
$$\begin{cases}
x_1=-4\sqrt 3\\
x_2=4\sqrt 3.
\end{cases}$$
Because $x_1$ and $x_2$ are the solutions of the equation, it means that
$$x-x_1=0\text{ and }x-x_2=0,$$
so the quadratic equations can be written:
$$a(x-x_1)(x-x_2)=0,\text{ where } a\text{ is any real number}.$$
In our case we have:
$$\begin{align*}
a\left(x-\left(-4\sqrt 3\right)\right)\left(x-4\sqrt 3\right)&=0\\
a\left(x+4\sqrt 3\right)\left(x-4\sqrt 3\right)&=0.
\end{align*}$$
For example take $a=1$:
$$\begin{align*}
\left(x+4\sqrt 3\right)\left(x-4\sqrt 3\right)&=0\\
x^2-(4\sqrt 3)^2&=0\\
x^2-48&=0.
\end{align*}$$
The equation is:
$$x^2-48=0.$$
Note: Any equation in the form $k(x^2-48)=0$, where $k$ is a real number, fits.