Correct option is (A) \(3, \frac{-3}2\)
Let the polynomial \(P(x)=2 x^2-3 x-9\).
\(P(x) =2 x^2-3 x-9\)
\(=2 x^2-6 x+3 x-9 \)
\(=2 x(x-3)+3(x-3)\)
\(=(x-3)(2 x+3)\)
The value of polynomial \(P(x) = 0\).
\(\Rightarrow (x-3)(2 x+3)=0\)
\(\Rightarrow(x-3)=0 \text { or } (2 x+3)=0\)
\(\Rightarrow x=3 \text { or } x=\frac{-3}{2}\)
Therefore, the zeroes of polynomial are 3 and \(\frac{-3}{2}\).