Correct. The area of the flag is $\displaystyle{2}\cdot{1}={2}\usepackage{cmbright}\Huge\$ . Adding a border of width x, the new area of the flag is $\displaystyle{\left({2}+{2}{x}\right)}{\left({1}+{2}{x}\right)}\usepackage{cmbright}\Huge\$ . The area of the border is given by the equation $\displaystyle{\left({2}+{2}{x}\right)}{\left({1}+{2}{x}\right)}-{\left({2}\right)}{\left({1}\right)}={10}\usepackage{cmbright}\Huge\$ , which can be rewritten as the trinomial $\displaystyle{4}{{x}}^{{2}}+{6}{x}-{10}={0}\usepackage{cmbright}\Huge\$ . Solving this trinomial yields $\displaystyle{x}={1}\usepackage{cmbright}\Huge\$ or $\displaystyle{x}=-\frac{{5}}{{2}}\usepackage{cmbright}\Huge\$ . Since $\displaystyle-\frac{{5}}{{2}}\usepackage{cmbright}\Huge\$ does not make sense in this context, the width of the border must be 1 foot.