Correct. Rewriting this equation as $\displaystyle\frac{{{3}{x}}}{{6}}-\frac{{{{x}}^{{2}}}}{{6}}={0}\usepackage{cmbright}\Huge\$ , you found that $\displaystyle{3}{x}-{{x}}^{{2}}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{{x}}^{{2}}-{3}{x}={x}{\left({x}-{3}\right)}={0}\usepackage{cmbright}\Huge\$ . This means $\displaystyle{x}={0}\usepackage{cmbright}\Huge\$ and $\displaystyle{x}={3}\usepackage{cmbright}\Huge\$ .