(2x2-x-3)/(x2-x-6)<0,
(2x-3)(x+1)/[(x-3)(x+2)]<0.
The quotient can only be negative is the numerator and denominator have different signs.
Therefore we need to solve for x when:
(1) (2x-3)(x+1)<0 and (x-3)(x+2)>0; or:
(2) (2x-3)(x+1)>0 and (x-3)(x+2)<0.
(1) (2x-3)(x+1)<0 when -1<x<1.5; (x-3)(x+2)>0 when x<-2 or x>3. So these conditions cannot be satisfied.
(2) (2x-3)(x+1)>0 when x<-1 or x>1.5; (x-3)(x+2)<0 when -2<x<3. This can be satisfied when -2<x<-1 or when 1.5<x<3.
CHECK:
When x=-1.5, the expression = (-6)(-0.5)/[(-4.5)(0.5)]=3/(-2.25)<0 OK.
When x=2, the expression = (1)(3)/[(-1)(4)]<0 OK.