Correct. If $\displaystyle\sqrt{{{x}}}-{13}={2}\usepackage{cmbright}\Huge\$ , then $\displaystyle\sqrt{{{x}}}={15}\usepackage{cmbright}\Huge\$ , so $\displaystyle{{\left(\sqrt{{{x}}}\right)}}^{{2}}={{15}}^{{2}}={225}\usepackage{cmbright}\Huge\$ , and therefore, $\displaystyle{x}={225}\usepackage{cmbright}\Huge\$ .