Correct. Factoring a 3 out of the original equation, you found that $\displaystyle{6}{{x}}^{{2}}+{3}{x}-{15}={3}{\left({2}{{x}}^{{2}}+{x}-{5}\right)}\usepackage{cmbright}\Huge\$ . Substituting the values $\displaystyle{a}={2},{b}={1},\usepackage{cmbright}\Huge\$ and $\displaystyle{c}=-{5}\usepackage{cmbright}\Huge\$ into the quadratic formula reveals that $\displaystyle{x}=\frac{{-{1}+\sqrt{{{41}}}}}{{4}}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}=\frac{{-{1}-\sqrt{{{41}}}}}{{4}}\usepackage{cmbright}\Huge\$ .