The equation can be rewritten as: $\displaystyle{{x}}^{{2}}-{x}-{1}={0}\usepackage{cmbright}\Huge\$

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

Therefore:

$\displaystyle{{b}}^{{2}}-{4}{a}{c}={{\left(-{1}\right)}}^{{2}}-{4}{\left({1}\right)}{\left(-{1}\right)}={1}+{4}={5}\usepackage{cmbright}\Huge\$

Because the discriminant is positive, there are two real solutions.