You may have correctly added 9 to both sides to get $\displaystyle{{x}}^{{2}}-{2}{x}+{9}={0}\usepackage{cmbright}\Huge\$ , and then substituted $\displaystyle{a}={1},{b}=-{2}\usepackage{cmbright}\Huge\$ , and $\displaystyle{c}={9}\usepackage{cmbright}\Huge\$ into the Quadratic Formula: $\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\usepackage{cmbright}\Huge\$

But you mistakenly thought that $\displaystyle-{b}\usepackage{cmbright}\Huge\$ was equal to $\displaystyle-{2}\usepackage{cmbright}\Huge\$ .