$\displaystyle{{\left(\sqrt{{{3}{a}+{1}}}\right)}}^{{2}}={{\left({a}-{3}\right)}}^{{2}}\usepackage{cmbright}\Huge\$

$\displaystyle{3}{a}+{1}={{a}}^{{2}}-{6}{a}+{9}\usepackage{cmbright}\Huge\$

$\displaystyle{0}={{a}}^{{2}}-{9}{a}+{8}\usepackage{cmbright}\Huge\$

$\displaystyle{0}={\left({a}-{1}\right)}{\left({a}-{8}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{a}={1}\usepackage{cmbright}\Huge\$ and $\displaystyle{a}={8}\usepackage{cmbright}\Huge\$

Check $\displaystyle{a}={1}\usepackage{cmbright}\Huge\$ : $\displaystyle\sqrt{{{3}{\left({1}\right)}+{1}}}={1}-{3}\usepackage{cmbright}\Huge\$ ? $\displaystyle{2}\ne-{2}\usepackage{cmbright}\Huge\$ extraneous solution
Check $\displaystyle{a}={8}\usepackage{cmbright}\Huge\$ : $\displaystyle\sqrt{{{3}{\left({8}\right)}+{1}}}={8}-{3}\usepackage{cmbright}\Huge\$ ? $\displaystyle\sqrt{{25}}={5}\usepackage{cmbright}\Huge\$ TRUE

Check $\displaystyle{a}={1}\usepackage{cmbright}\Huge\$ :	$\displaystyle\sqrt{{{3}{\left({1}\right)}+{1}}}={1}-{3}\usepackage{cmbright}\Huge\$ ?	$\displaystyle{2}\ne-{2}\usepackage{cmbright}\Huge\$	extraneous solution
Check $\displaystyle{a}={8}\usepackage{cmbright}\Huge\$ :	$\displaystyle\sqrt{{{3}{\left({8}\right)}+{1}}}={8}-{3}\usepackage{cmbright}\Huge\$ ?	$\displaystyle\sqrt{{25}}={5}\usepackage{cmbright}\Huge\$	TRUE