The equation can be rewritten 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\$ .

Therefore:

$\displaystyle{{b}}^{{2}}-{4}{a}{c}={{\left({4}\right)}}^{{2}}-{4}{\left({3}\right)}{\left({5}\right)}={16}-{60}=-{44}\usepackage{cmbright}\Huge\$

Because the discriminant is $\displaystyle-{44}\usepackage{cmbright}\Huge\$ , there are two complex (nonreal) solutions.