Correct. If $\displaystyle\sqrt{{{x}}}-{4}={3}\sqrt{{{x}}}-{8}\usepackage{cmbright}\Huge\$ , then $\displaystyle{4}={2}\sqrt{{{x}}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{2}=\sqrt{{{x}}}\usepackage{cmbright}\Huge\$ . Therefore, $\displaystyle{{2}}^{{2}}={{\left(\sqrt{{{x}}}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={4}\usepackage{cmbright}\Huge\$ .