Correct. Setting the equations equal, you find $\displaystyle{3}{x}+{4}={2}{{x}}^{{2}}+{2}\usepackage{cmbright}\Huge\$ , which can be rewritten $\displaystyle{2}{{x}}^{{2}}-{3}{x}-{2}={0}\usepackage{cmbright}\Huge\$ . This can be factored as $\displaystyle{\left({2}{x}+{1}\right)}{\left({x}-{2}\right)}={0}\usepackage{cmbright}\Huge\$ , so there are two solutions.