Correct. Rewrite the equation so that the right side is 0: $\displaystyle{3}{{\left({x}+{1}\right)}}^{{2}}-{8}{x}+{1}={0}\usepackage{cmbright}\Huge\$

Square the binomial: $\displaystyle{3}{\left({{x}}^{{2}}+{2}{x}+{1}\right)}-{8}{x}+{1}={0}\usepackage{cmbright}\Huge\$

Multiply out: $\displaystyle{3}{{x}}^{{2}}+{6}{x}+{3}-{8}{x}+{1}={0}\usepackage{cmbright}\Huge\$

Combine like terms: $\displaystyle{3}{{x}}^{{2}}-{2}{x}+{4}={0}\usepackage{cmbright}\Huge\$

The value of $\displaystyle{b}=-{2}\usepackage{cmbright}\Huge\$ .