Answer
$\displaystyle \frac{(4x+1)(2x-3)}{(4x-1)(3x-1)}$
Work Step by Step
$\displaystyle \frac{\frac{9x^{2}+3x-2}{12x^{2}+5x-2}}{\frac{9x^{2}-6x+1}{8x^{2}-10x-3}}=\frac{9x^{2}+3x-2}{12x^{2}+5x-2}\div\frac{9x^{2}-6x+1}{8x^{2}-10x-3}$
$=\displaystyle \frac{9x^{2}+3x-2}{12x^{2}+5x-2}\cdot\frac{8x^{2}-10x-3}{9x^{2}-6x+1}=...$
Factor each trinomial
(for $ax^{2}+bx+c$ we search for factors of ac that add up to b.)
$9x^{2}+3x-2=9x^{2}+6x-3x-2$
$=3x(3x+2)+-(3x+2)$
$=(3x-1)(3x+2)$
$12x^{2}+5x-2=12x^{2}+8x-3x-2$
$=4x(3x+2)-(3x+2)$
$=(4x-1)(3x+2)$
$8x^{2}-10x-3=8x^{2}-12x+2x-3$
$=4x(2x-3)+(2x-3)$
$=(4x+1)(2x-3)$
$9x^{2}-6x+1=9x^{2}-3x-3x+1$
$=3x(3x-1)-(3x-1)$
$=(3x-1)(3x-1)$
$...=\displaystyle \frac{(3x-1)(3x+2)}{(4x-1)(3x+2)}\cdot\frac{(4x+1)(2x-3)}{(3x-1)(3x-1)}=$
Common factors $(3x-1)$ and $(3x+2)$ cancel,
$=\displaystyle \frac{(4x+1)(2x-3)}{(4x-1)(3x-1)}$
