Correct. Setting the equations equal, you have $\displaystyle-{4}{{x}}^{{2}}-{8}{x}=-{3}{{x}}^{{2}}+{12}\usepackage{cmbright}\Huge\$ , which can be rewritten as $\displaystyle{{x}}^{{2}}+{8}{x}+{12}={0}\usepackage{cmbright}\Huge\$ . This can be factored as $\displaystyle{\left({x}+{6}\right)}{\left({x}+{2}\right)}={0}\usepackage{cmbright}\Huge\$ , so there are two solutions.