The equation can be rewritten as: $\displaystyle{2}{{x}}^{{2}}-{x}+{4}={0}\usepackage{cmbright}\Huge\$

In this equation, $\displaystyle{a}={2},{b}=-{1}\usepackage{cmbright}\Huge\$ , and $\displaystyle{c}={4}\usepackage{cmbright}\Huge\$ .

Therefore:

$\displaystyle{{b}}^{{2}}-{4}{a}{c}={{\left(-{1}\right)}}^{{2}}-{4}{\left({2}\right)}{\left({4}\right)}={1}-{32}=-{31}\usepackage{cmbright}\Huge\$

Because the discriminant is negative, there are two complex (nonreal) solutions.