Correct. Substituting $\displaystyle{2}{{x}}^{{2}}-{20}\usepackage{cmbright}\Huge\$ in for y in the second equation, you find $\displaystyle{2}{{x}}^{{2}}-{20}={12}{x}-{6}\usepackage{cmbright}\Huge\$ . This can be rewritten as $\displaystyle{2}{{x}}^{{2}}-{12}{x}-{14}={0}\usepackage{cmbright}\Huge\$ , and then factored as $\displaystyle{2}{\left({x}+{1}\right)}{\left({x}-{7}\right)}={0}\usepackage{cmbright}\Huge\$ . Once you find that $\displaystyle{x}=-{1}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}={7},\usepackage{cmbright}\Huge\$ you can substitute these values back into the original equations to find the corresponding y-values.