Correct. In this equation, $\displaystyle{a}={4},{b}={12}\usepackage{cmbright}\Huge\$ , and $\displaystyle{c}={9}\usepackage{cmbright}\Huge\$ . Therefore,

$\displaystyle{{b}}^{{2}}-{4}{a}{c}={{12}}^{{2}}-{4}{\left({4}\right)}{\left({9}\right)}={144}-{144}={0}\usepackage{cmbright}\Huge\$

Since the discriminant is 0, there is one real solution.