Answer
$= \frac{(3c-5)(c-5)}{(2c+3)(c+4)}$
Work Step by Step
$\frac{c^{2}-c-20}{c^{2}+6c+8} \div \frac{2c^{2}+11c+12}{3c^{2}+c-10}$
$= \frac{c^{2}-c-20}{c^{2}+6c+8} \times \frac{3c^{2}+c-10}{2c^{2}+11c+12}$
$= \frac{(c-5)(c+4)}{(c+4)(c+2)} \times \frac{3c^{2}-5c+6c-10}{2c^{2}+8c+3c+12}$
$= \frac{(c-5)}{(c+2)} \times \frac{c(3c-5)+2(3c-5)}{2c(c+4)+3(c+4)}$
$= \frac{(c-5)}{(c+2)} \times \frac{(c+2)(3c-5)}{(2c+3)(c+4)}$
$= \frac{(c-5)}{1} \times \frac{(3c-5)}{(2c+3)(c+4)}$
$= \frac{(3c-5)(c-5)}{(2c+3)(c+4)}$