The question is simplfy: $\displaystyle \frac{t^2-9}{6-2t} \times \frac{6}{t-3} $

i simply till $\displaystyle \frac{3(t-3)}{3-t)} $ then i mulitpy by -1 for both denominator and numerator so i get -3 in the end. is this correct? the ans given is$\displaystyle \frac{3(t-3)}{3-t)}$ . but i can get -3. which shld be the correct ans?