Incorrect. Rewrite in standard form by subtracting $\displaystyle{4}{x}+{15}\usepackage{cmbright}\Huge\$ from both sides:

$\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\$ .