
L'équation admet deux solutions distinctes uniquement si $\Delta>0$. Or,
\begin{align*}
\Delta & = b^2-4ac\\
& = (-4m)^2 - 4\times(4m+1)\times(m-3)\\
& = 16m^2-4(4m+1)(m-3)\\
& = 16m^2-4(4m^2-12m+m-3)\\
& = 16m^2 - 4(4m^2-11m-3)\\
& = 16m^2-16m^2+44m+12\\
& = 44m+12.
\end{align*}

Ainsi,
\begin{align*}
\Delta > 0 & \iff 44m+12>0\\
& \iff 44m > -12\\
& \iff m > \dfrac{-12}{44}\\
& \iff m > -\dfrac{3}{11}.
\end{align*}

\underline{Par conséquent, l'équation admet deux solutions distinctes pour $m>-\dfrac{3}{11}$.}
