Correct. This equation can be factored without completing the square: $\displaystyle{4}{{x}}^{{2}}+{16}{x}-{48}={4}{\left({x}+{6}\right)}{\left({x}-{2}\right)}\usepackage{cmbright}\Huge\$ . The solutions are $\displaystyle{x}=-{6}\usepackage{cmbright}\Huge\$ and $\displaystyle{x}={2}\usepackage{cmbright}\Huge\$ .