[tex]x_1+x_2+x_3= -\frac{b}{a} 0\\
x_1x_2+x_1x_3+x_2x_3= \frac{c}{a} =m\\
x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=-2m\\
Deoarece~m\ \textgreater \ 0~atunci~x_1^2+x_2^2+x_3^2\ \textless \ 0.[/tex]
Ultima inegalitate ne arata ca f nu are toate radacinile reale. In concluzie, va avea doua radacini complexe(una conjugata celeilalte, adica vor avea modulele egale) si o radacina reala.
