Correct. Rewriting the equation as $\displaystyle{{x}}^{{2}}+{6}{x}-{1}={0}\usepackage{cmbright}\Huge\$ and then substituting the values $\displaystyle{a}={1},{b}={6},\usepackage{cmbright}\Huge\$ and $\displaystyle{c}=-{1}\usepackage{cmbright}\Huge\$ into the quadratic formula, you found that $\displaystyle{x}=\frac{{-{6}\pm{2}\sqrt{{{10}}}}}{{2}}\usepackage{cmbright}\Huge\$ $\displaystyle=\frac{{2}}{{2}}\cdot{\left(\frac{{-{3}\pm\sqrt{{{10}}}}}{{1}}\right)}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}=-{3}+\sqrt{{{10}}}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}=-{3}-\sqrt{{{10}}}\usepackage{cmbright}\Huge\$ .