First rewrite as $\displaystyle{{x}}^{{2}}-{2}{x}-{15}={0}\usepackage{cmbright}\Huge\$ . Then use the Quadratic Formula, $\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\usepackage{cmbright}\Huge\$ , with $\displaystyle{a}={1},{b}=-{2}\usepackage{cmbright}\Huge\$ , and $\displaystyle{c}=-{15}\usepackage{cmbright}\Huge\$ :

$\displaystyle{x}=\frac{{-{\left(-{2}\right)}\pm\sqrt{{{{\left(-{2}\right)}}^{{2}}-{4}{\left({1}\right)}{\left(-{15}\right)}}}}}{{{2}\cdot{1}}}=\frac{{{2}\pm\sqrt{{{4}+{60}}}}}{{2}}=\frac{{{2}\pm\sqrt{{64}}}}{{2}}\usepackage{cmbright}\Huge\$

$\displaystyle{x}=\frac{{{2}\pm{8}}}{{2}}\usepackage{cmbright}\Huge\$

$\displaystyle{x}=\frac{{{2}+{8}}}{{2}}=\frac{{10}}{{2}}={5}\usepackage{cmbright}\Huge\$

$\displaystyle{x}=\frac{{{2}-{8}}}{{2}}=\frac{{-{6}}}{{2}}=-{3}\usepackage{cmbright}\Huge\$