Correct. Substituting the values $\displaystyle{a}={3},{b}={5},\usepackage{cmbright}\Huge\$ and $\displaystyle{c}={0}\usepackage{cmbright}\Huge\$ into the quadratic formula, you found $\displaystyle{x}=\frac{{-{5}\pm\sqrt{{{{5}}^{{2}}-{4}{\left({3}\right)}{\left({0}\right)}}}}}{{{2}{\left({3}\right)}}}=\frac{{-{5}\pm\sqrt{{{25}}}}}{{6}}\usepackage{cmbright}\Huge\$ so $\displaystyle{x}={0}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}=-\frac{{5}}{{3}}\usepackage{cmbright}\Huge\$ . Notice that this equation can also be factored without using the quadratic formula: $\displaystyle{3}{{x}}^{{2}}+{5}{x}={x}{\left({3}{x}+{5}\right)}.\usepackage{cmbright}\Huge\$