Question 16799
Do you mean:

{{{(6y^2 -y -2) / ((2y+1) /(3y-2))}}}


If so, then you have two main problems:  First you must factor the numerator, and second, you must invert the denominator and multiply.


{{{ ((3y-2)(2y+1)) * ((3y-2)/(2y+1)) }}}


The (2y+1) factors divide out, and your final answer is {{{(3y-2)^2}}}.


R%2 at SCC