#### Answer

$\dfrac{t-4}{t+3}$

#### Work Step by Step

Adding the numerators and copying the denominator, the given expression, $
\dfrac{t^2-5t}{t^2+6t+9}+\dfrac{4t-12}{t^2+6t+9}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{t^2-5t+4t-12}{t^2+6t+9}
\\\\=
\dfrac{t^2-t-12}{t^2+6t+9}
\\\\=
\dfrac{(t-4)(t+3)}{(t+3)(t+3)}
\\\\=
\dfrac{(t-4)(\cancel{t+3})}{(t+3)(\cancel{t+3})}
\\\\=
\dfrac{t-4}{t+3}
.\end{array}