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

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

Therefore:

$\displaystyle{{b}}^{{2}}-{4}{a}{c}={{\left(-{4}\right)}}^{{2}}-{4}{\left({2}\right)}{\left({2}\right)}={16}-{16}={0}\usepackage{cmbright}\Huge\$

Because the discriminant is 0, there is one real solution.