Correct. $\displaystyle{{b}}^{{2}}+{2}{b}-{8}={\left({b}+{4}\right)}{\left({b}-{2}\right)}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{b}=-{4}\usepackage{cmbright}\Huge\$ or $\displaystyle{b}={2}\usepackage{cmbright}\Huge\$ .