Correct. Since $\displaystyle{64}={{\left(-{8}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ and $\displaystyle-{16}{x}={2}\cdot{x}\cdot{\left(-{8}\right)}\usepackage{cmbright}\Huge\$ , the quadratic expression is a perfect square trinomial. Therefore,

$\displaystyle{\left({x}-{8}\right)}{\left({x}-{8}\right)}={0}\usepackage{cmbright}\Huge\$

$\displaystyle{x}-{8}={0}\usepackage{cmbright}\Huge\$

$\displaystyle{x}={8}\usepackage{cmbright}\Huge\$