A trinomial in the form $\displaystyle{{r}}^{{2}}-{2}{r}{s}+{{s}}^{{2}}\usepackage{cmbright}\Huge\$ can be factored as $\displaystyle{{\left({r}-{s}\right)}}^{{2}}\usepackage{cmbright}\Huge\$ . So the left side is:

$\displaystyle{25}{{x}}^{{2}}-{20}{x}+{4}={{\left({5}{x}\right)}}^{{2}}-{2}{\left({5}{x}\right)}{\left({2}\right)}+{{2}}^{{2}}={{\left({5}{x}-{2}\right)}}^{{2}}\usepackage{cmbright}\Huge\$

The equation can be written as: $\displaystyle{{\left({5}{x}-{2}\right)}}^{{2}}={45}\usepackage{cmbright}\Huge\$

Use the Square Root Property: $\displaystyle{5}{x}-{2}=\pm\sqrt{{45}}=\pm\sqrt{{{9}\cdot{5}}}=\pm{3}\sqrt{{5}}\usepackage{cmbright}\Huge\$

Solve for x: $\displaystyle{5}{x}={2}\pm{3}\sqrt{{5}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}=\frac{{2}}{{5}}\pm\frac{{3}}{{5}}\sqrt{{5}}\usepackage{cmbright}\Huge\$