Solution
(A)
$$6x^2+x-12=0$$
$$6x^2+9x-8x-12=0$$
$$3x(2x+3)-4(2x+3)=0$$
$$(3x-4)(2x+3)=0$$
$$x=\dfrac{4}{3}\ or\ x=-\dfrac{3}{2}$$
(A) - (IV)
(D)
$$2x^2-5x-3=0$$
$$2x^2-6x+x-3=0$$
$$2x(x-3)+1(x-3)=0$$
$$(2x+1)(x-3)=0$$
$$x=\dfrac{-1}{2}\ or\ x=3$$
(D) - (III)
Thus, the correct answer is option B.
NOTE:- Equation (B) and Equation (C) were incorrectly typed in the actual examination as well by NTA.
