Correct. Finding a common denominator and setting the numerators equal, you find $\displaystyle{{x}}^{{2}}=-{5}{x}\usepackage{cmbright}\Huge\$ , so $\displaystyle{{x}}^{{2}}+{5}{x}={x}{\left({x}+{5}\right)}={0}\usepackage{cmbright}\Huge\$ , meaning that $\displaystyle{x}={0}\usepackage{cmbright}\Huge\$ and -5. However, the domain excludes $\displaystyle{x}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}=-{5}\usepackage{cmbright}\Huge\$ is the correct solution.