F are maxim, daca m<0 ("parabola nu tine apa")
[tex]f_{max}=f(-\dfrac {b}{2a})=f(\dfrac4m)=m\cdot\dfrac{16}{m^2}-\dfrac{32}{m}-3\Rightarrow \dfrac{16}{m}-\dfrac{32}{m}-3=5[/tex]
Aducem la acelasi numitor, si avem
[tex]-16=8m\Rightarrow m=-2.\ \ \ (-2<0)[/tex]