These can be factorised to make things simpler.
2x²-5x+3=(x-1)(2x-3); x²-4=(x-2)(x+2); 2x²+7x+3=(2x+1)(x+3); x²+x-6=(x+3)(x-2).
Now rewrite the expression:
[(x-1)(2x-3)][(x+3)(x-2)]/([(x-2)(x+2)][(2x+1)(x+3)])=
[(x-1)(2x-3)]/([(x+2)(2x+1)])=
(2x²-5x+3)/(2x²+5x+2).