Correct. $\displaystyle{y}={2}{{x}}^{{2}}-{20}{x}={2}{\left({{x}}^{{2}}-{10}{x}\right)}\usepackage{cmbright}\Huge\$ . Completing the square, you found $\displaystyle{y}+{2}{\left({25}\right)}={2}{\left({{x}}^{{2}}-{10}{x}+{25}\right)}\usepackage{cmbright}\Huge\$ . Factoring the trinomial, this can be rewritten as $\displaystyle{y}={2}{{\left({x}-{5}\right)}}^{{2}}-{50}\usepackage{cmbright}\Huge\$ . This equation is now in vertex form, $\displaystyle{y}={a}{{\left({x}-{h}\right)}}^{{2}}+{k}\usepackage{cmbright}\Huge\$ , and you can find the vertex, $\displaystyle{\left({h},{k}\right)}\usepackage{cmbright}\Huge\$ . The coordinates of the vertex are $\displaystyle{\left({5},-{50}\right)}\usepackage{cmbright}\Huge\$ .