You would add 64 to $\displaystyle{{x}}^{{2}}+{16}{x}\usepackage{cmbright}\Huge\$ to make it $\displaystyle{{\left({x}+{8}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ .

If you start with $\displaystyle{{x}}^{{2}}+{b}{x}\usepackage{cmbright}\Huge\$ , add the number $\displaystyle{{\left(\frac{{b}}{{2}}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ to complete the square.

The resulting trinomial will be: $\displaystyle{{x}}^{{2}}+{b}{x}+{{\left(\frac{{b}}{{2}}\right)}}^{{2}}={{\left({x}+\frac{{b}}{{2}}\right)}}^{{2}}\usepackage{cmbright}\Huge\$

Here $\displaystyle{b}={8}\usepackage{cmbright}\Huge\$ .