Subtract 3 from both sides: $\displaystyle\sqrt{{x}}=-{2}\usepackage{cmbright}\Huge\$

Square both sides and simplify: $\displaystyle{{\left(\sqrt{{x}}\right)}}^{{2}}={{\left(-{2}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={4}\usepackage{cmbright}\Huge\$

Check: $\displaystyle\sqrt{{4}}+{3}={1}\usepackage{cmbright}\Huge\$ ? $\displaystyle{2}+{3}\ne{1}\usepackage{cmbright}\Huge\$

So $\displaystyle{x}={4}\usepackage{cmbright}\Huge\$ is an extraneous solution.