Đáp án:
$\begin{array}{l}
{x^2} - \left( {2m + 1} \right)x + {m^2} = 0\\
\Rightarrow \Delta = {\left( {2m + 1} \right)^2} - 4{m^2} = 4{m^2} + 4m + 1 - 4{m^2}\\
= 4m + 1\left( {m \ge \frac{{ - 1}}{4}} \right)\\
\Rightarrow x = \frac{{ - b \pm \sqrt \Delta }}{2} = \frac{{2m + 1 \pm \sqrt {4m + 1} }}{2}
\end{array}$