Answer
$\frac{(2n+1)}{(3n+4)}$
Work Step by Step
Since both the numerator and the denominator consist of trinomials, we can use the rules for factoring trinomials to factor these trinomails and cancel out the common factors:
$\frac{2n^{2}-7n-4}{3n^{2}-8n-16}$
=$\frac{2n^{2}+1n-8n-4}{3n^{2}+4n-12n-16}$
=$\frac{n(2n+1)-4(2n+1)}{n(3n+4)-4(3n+4)}$
=$\frac{(2n+1)(n-4)}{(3n+4)(n-4)}$
=$\frac{(2n+1)}{(3n+4)}$