Correct. $\displaystyle{{a}}^{{2}}+{6}{a}+{5}={\left({a}+{5}\right)}{\left({a}+{1}\right)}={0},\usepackage{cmbright}\Huge\$ so $\displaystyle{a}=-{5}\usepackage{cmbright}\Huge\$ or $\displaystyle{a}=-{1}\usepackage{cmbright}\Huge\$ .