Rewrite in standard form: $\displaystyle{{x}}^{{2}}-{6}{x}+{4}={0}\usepackage{cmbright}\Huge\$ . So $\displaystyle{a}={1},{b}=-{6}\usepackage{cmbright}\Huge\$ , and $\displaystyle{c}={4}\usepackage{cmbright}\Huge\$ .

Use the Quadratic Formula: $\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\usepackage{cmbright}\Huge\$

$\displaystyle{x}=\frac{{-{\left(-{6}\right)}\pm\sqrt{{{{\left(-{6}\right)}}^{{2}}-{4}{\left({1}\right)}{\left({4}\right)}}}}}{{{2}\cdot{1}}}=\frac{{{6}\pm\sqrt{{{36}-{16}}}}}{{2}}=\frac{{{6}\pm\sqrt{{{20}}}}}{{2}}\usepackage{cmbright}\Huge\$

$\displaystyle=\frac{{{6}\pm\sqrt{{4}}\sqrt{{5}}}}{{2}}=\frac{{{6}\pm{2}\sqrt{{5}}}}{{2}}={3}\pm\sqrt{{5}}\usepackage{cmbright}\Huge\$