Correct. Rewrite the equation as $\displaystyle{4}{{x}}^{{2}}-{7}{x}+{3}={0}\usepackage{cmbright}\Huge\$ . Then solve: $\displaystyle{4}{{x}}^{{2}}-{7}{x}+{3}={\left({4}{x}-{3}\right)}{\left({x}-{1}\right)}={0}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}=\frac{{3}}{{4}}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}={1}\usepackage{cmbright}\Huge\$ .