Correct. Because $\displaystyle{a}={2}\usepackage{cmbright}\Huge\$ , the first term will be $\displaystyle{2}{{x}}^{{2}}\usepackage{cmbright}\Huge\$ . If we rewrite the equation with the right side equal to 0, the left side will have the term $\displaystyle{2}{{x}}^{{2}}\usepackage{cmbright}\Huge\$ . To get 0 on the right side, add the opposite of $\displaystyle-{2}{{x}}^{{2}}+{3}{x}+{1}\usepackage{cmbright}\Huge\$ to both sides:

$\displaystyle{8}{x}+{4}+{2}{{x}}^{{2}}-{3}{x}-{1}=-{2}{{x}}^{{2}}+{3}{x}+{1}+{2}{{x}}^{{2}}-{3}{x}-{1}\usepackage{cmbright}\Huge\$

$\displaystyle{2}{{x}}^{{2}}+{5}{x}+{3}={0}\usepackage{cmbright}\Huge\$

The value of $\displaystyle{b}={5}\usepackage{cmbright}\Huge\$ , the coefficient of the x term.