Option A: $3, \dfrac{-3}{2}$
Factorising: $2x^2 - 3x - 9 = 2x^2 - 6x + 3x - 9 = 2x(x-3) + 3(x-3) = (x-3)(2x+3)$
Zeroes: $x = 3$ or $x = -\dfrac{3}{2}$
Source: Chapter 2, Section 2.3
Split the middle term $-3x$ as $-6x + 3x$ (product = $2 \times (-9) = -18$, sum = $-3$). After factorising, set each factor to zero. Examiners expect the factorisation step shown — don't just state the answer for a calculation-based MCQ if working is expected.