Incorrect. Remember that the quadratic formula is $\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\usepackage{cmbright}\Huge\$ . Notice that the term $\displaystyle{2}{a}\usepackage{cmbright}\Huge\$ must evaluate to 16. What value must a be?