Correct. $\displaystyle{{t}}^{{2}}+{3}{t}+{2}={\left({t}+{2}\right)}{\left({t}+{1}\right)}={0}\usepackage{cmbright}\Huge\$ ; $\displaystyle{t}=-{2}\usepackage{cmbright}\Huge\$ or $\displaystyle{t}=-{1}\usepackage{cmbright}\Huge\$