Answer
$15x^2-4x-3=0$
Work Step by Step
We are given the solutions of a quadratic equation:
$$\begin{cases}
x_1=-\dfrac{1}{3}\\
x_2=\dfrac{3}{5}.
\end{cases}$$
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:
$$\begin{align*}
a\left(x-\left(-\dfrac{1}{3}\right)\right)\left(x-\dfrac{3}{5}\right)&=0\\
a\left(x+\dfrac{1}{3}\right)\left(x-\dfrac{3}{5}\right)&=0.
\end{align*}$$
For example take $a=15$:
$$\begin{align*}
15\left(x+\dfrac{1}{3}\right)\left(x-\dfrac{3}{5}\right)&=0\\
(3x+1)(5x-3)&=0\\
15x^2-4x-3&=0.
\end{align*}$$
The equation is:
$$15x^2-4x-3=0.$$
Note: Any equation in the form $k(15x^2-4x-3)=0$, where $k$ is a real number, fits.