A quadratic polynomial whose zeros are

(d) $10 x^{2}-x-3$

Here, the zeroes are $\frac{3}{5}$ and $\frac{-1}{2}$.

Let $\alpha=\frac{3}{5}$ and $\beta=\frac{-1}{2}$

So, sum of the zeroes, $\alpha+\beta=\frac{3}{5}+\left(\frac{-1}{2}\right)=\frac{1}{10}$

Also, product of the zeroes, $\alpha \beta=\frac{3}{5} \times\left(\frac{-1}{2}\right)=\frac{-3}{10}$

The polynomial will be $x^{2}-(\alpha+\beta) x+\alpha \beta$.

$\therefore$ The required polynomial is $x^{2}-\frac{1}{10} x-\frac{3}{10}$.

Multiply by 10, we get

$10 x^{2}-x-3$