Correct. A common denominator is $\displaystyle{\left({t}+{3}\right)}{\left({t}-{2}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle\frac{{{2}{t}+{1}}}{{{t}+{3}}}-\frac{{{t}-{4}}}{{{t}-{2}}}=\frac{{{\left({t}-{2}\right)}{\left({2}{t}+{1}\right)}-{\left({t}-{4}\right)}{\left({t}+{3}\right)}}}{{{\left({t}+{3}\right)}{\left({t}-{2}\right)}}}=\frac{{{{t}}^{{2}}-{2}{t}+{10}}}{{{{t}}^{{2}}+{t}-{6}}}\usepackage{cmbright}\Huge\$ . The domain excludes the values $\displaystyle{t}=-{3}\usepackage{cmbright}\Huge\$ and $\displaystyle{t}={2}\usepackage{cmbright}\Huge\$ because those values would lead to a denominator of 0.