You may have miscalculated the discriminant $\displaystyle{{b}}^{{2}}-{4}{a}{c}\usepackage{cmbright}\Huge\$ . If the discriminant were 0, there would be one real solution. You need to rewrite the equation as $\displaystyle{3}{{x}}^{{2}}+{4}{x}+{5}={0}\usepackage{cmbright}\Huge\$ . In this equation, $\displaystyle{a}={3},{b}={4}\usepackage{cmbright}\Huge\$ , and $\displaystyle{c}={5}\usepackage{cmbright}\Huge\$ .