$\displaystyle{{\left(\sqrt{{{3}{a}-{5}}}\right)}}^{{2}}={{\left({a}-{3}\right)}}^{{2}}\usepackage{cmbright}\Huge\$

$\displaystyle{3}{a}-{5}={{a}}^{{2}}-{6}{a}+{9}\usepackage{cmbright}\Huge\$

$\displaystyle{0}={{a}}^{{2}}-{9}{a}+{14}\usepackage{cmbright}\Huge\$

$\displaystyle{0}={\left({a}-{2}\right)}{\left({a}-{7}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{a}={2}\usepackage{cmbright}\Huge\$ and $\displaystyle{a}={7}\usepackage{cmbright}\Huge\$

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

Check $\displaystyle{a}={7}\usepackage{cmbright}\Huge\$ : $\displaystyle\sqrt{{{3}{\left({7}\right)}-{5}}}={7}-{3}\usepackage{cmbright}\Huge\$ ? $\displaystyle\sqrt{{16}}={4}\usepackage{cmbright}\Huge\$ TRUE