The zeros of the polynomial p(x)

The given polynomial is $p(x)=2 x^{2}+5 x-3$.

Putting $x=\frac{1}{2}$ in $p(x)$, we get

$p\left(\frac{1}{2}\right)=2 \times\left(\frac{1}{2}\right)^{2}+5 \times \frac{1}{2}-3=\frac{1}{2}+\frac{5}{2}-3=3-3=0$

Therefore, $x=\frac{1}{2}$ is a zero of the polynomial $p(x)$.

Putting x = –3 in p(x), we get

$p(-3)=2 \times(-3)^{2}+5 \times(-3)-3=18-15-3=0$

Therefore, x = –3 is a zero of the polynomial p(x).

Thus, $\frac{1}{2}$ and $-3$ are the zeroes of the given polynomial $p(x)$.

Hence, the correct answer is option (b).