Correct. The equation $\displaystyle{3}={2}{{x}}^{{2}}+{4}{x}\usepackage{cmbright}\Huge\$ can be rewritten as $\displaystyle{0}={2}{{x}}^{{2}}+{4}{x}-{3}\usepackage{cmbright}\Huge\$ , so $\displaystyle{a}={2},{b}={4},\usepackage{cmbright}\Huge\$ and $\displaystyle{c}=-{3}\usepackage{cmbright}\Huge\$ . Applying the quadratic formula, you have $\displaystyle{x}=\frac{{-{4}\pm\sqrt{{{{\left({4}\right)}}^{{2}}-{4}{\left({2}\right)}{\left(-{3}\right)}}}}}{{{2}{\left({2}\right)}}}={x}=\frac{{-{4}\pm\sqrt{{{16}+{24}}}}}{{{4}}}\usepackage{cmbright}\Huge\$ . (Notice that this formula would yield the same result for x had you rewritten the original equation as $\displaystyle-{2}{{x}}^{{2}}-{4}{x}+{3}={0}\usepackage{cmbright}\Huge\$ .)