Square both sides: $\displaystyle{{\left(\sqrt{{{x}+{1}}}\right)}}^{{2}}={{\left({x}-{5}\right)}}^{{2}}\usepackage{cmbright}\Huge\$

Simplify and multiply: $\displaystyle{x}+{1}={{x}}^{{2}}-{10}{x}+{25}\usepackage{cmbright}\Huge\$

Subtract $\displaystyle{x}+{1}\usepackage{cmbright}\Huge\$ from both sides: $\displaystyle{0}={{x}}^{{2}}-{11}{x}+{24}\usepackage{cmbright}\Huge\$

Factor and solve: $\displaystyle{0}={\left({x}-{3}\right)}{\left({x}-{8}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={3}\usepackage{cmbright}\Huge\$ and $\displaystyle{x}={8}\usepackage{cmbright}\Huge\$

Check $\displaystyle{x}={3}\usepackage{cmbright}\Huge\$ : $\displaystyle\sqrt{{{3}+{1}}}={3}-{5}\usepackage{cmbright}\Huge\$ ? $\displaystyle\sqrt{{4}}\ne-{2}\usepackage{cmbright}\Huge\$ extraneous solution
Check $\displaystyle{x}={8}\usepackage{cmbright}\Huge\$ : $\displaystyle\sqrt{{{8}+{1}}}={8}-{5}\usepackage{cmbright}\Huge\$ ? $\displaystyle\sqrt{{9}}={3}\usepackage{cmbright}\Huge\$ TRUE

Check $\displaystyle{x}={3}\usepackage{cmbright}\Huge\$ :	$\displaystyle\sqrt{{{3}+{1}}}={3}-{5}\usepackage{cmbright}\Huge\$ ?	$\displaystyle\sqrt{{4}}\ne-{2}\usepackage{cmbright}\Huge\$	extraneous solution
Check $\displaystyle{x}={8}\usepackage{cmbright}\Huge\$ :	$\displaystyle\sqrt{{{8}+{1}}}={8}-{5}\usepackage{cmbright}\Huge\$ ?	$\displaystyle\sqrt{{9}}={3}\usepackage{cmbright}\Huge\$	TRUE