Correct. $\displaystyle{2}{{x}}^{{2}}-{7}{x}={4}\usepackage{cmbright}\Huge\$ can be rewritten as $\displaystyle{2}{{x}}^{{2}}-{7}{x}-{4}={0}\usepackage{cmbright}\Huge\$ . Factoring the polynomial, you find $\displaystyle{2}{{x}}^{{2}}-{7}{x}-{4}={\left({2}{x}+{1}\right)}{\left({x}-{4}\right)}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}=-\frac{{1}}{{2}}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}={4}\usepackage{cmbright}\Huge\$ .