$\displaystyle{{\left(\sqrt{{{2}{a}-{5}}}\right)}}^{{2}}={{\left({a}-{4}\right)}}^{{2}}\usepackage{cmbright}\Huge\$

$\displaystyle{2}{a}-{5}={{a}}^{{2}}-{8}{a}+{16}\usepackage{cmbright}\Huge\$

$\displaystyle{0}={{a}}^{{2}}-{10}{a}+{21}\usepackage{cmbright}\Huge\$

$\displaystyle{0}={\left({a}-{3}\right)}{\left({a}-{7}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{a}={3}\usepackage{cmbright}\Huge\$ and $\displaystyle{a}={7}\usepackage{cmbright}\Huge\$

Check $\displaystyle{a}={3}\usepackage{cmbright}\Huge\$ :
$\displaystyle\sqrt{{{2}{\left({3}\right)}-{5}}}={3}-{4}\usepackage{cmbright}\Huge\$ ?
$\displaystyle\sqrt{{{1}}}\ne-{1}\usepackage{cmbright}\Huge\$
extraneous solution
Check $\displaystyle{a}={7}\usepackage{cmbright}\Huge\$ :
$\displaystyle\sqrt{{{2}{\left({7}\right)}-{5}}}={7}-{4}\usepackage{cmbright}\Huge\$ ? $\displaystyle\sqrt{{9}}={3}\usepackage{cmbright}\Huge\$
TRUE