Correct. Rewriting the equation as $\displaystyle{{x}}^{{2}}-{9}{x}-{10}={0}\usepackage{cmbright}\Huge\$ , you find that it factors to $\displaystyle{\left({x}-{10}\right)}{\left({x}+{1}\right)}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={10}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}=-{1}\usepackage{cmbright}\Huge\$ .