Resposta :
[tex]\Delta >0\Rightarrow \exists x_1, x_2, \ x_1\neq x_2:ax^2+bx+c=0,\ x=x_1 \ ou \ x=x_2[/tex]
Prova:
[tex]{\Delta >0,\Delta <0,\Delta =0}\\\\x=\frac{-b\mp \sqrt{\Delta }}{2a}\\\Delta =0\Rightarrow x=\frac{-b\mp \sqrt{0}}{2a}\Rightarrow x=\frac{-b}{2a}\\\\\Delta <0\Rightarrow x=\frac{-b\mp \sqrt{-\Delta }}{2a}, \sqrt{-\Delta}=m:m\notin \mathbb{R}\Rightarrow x\notin \mathbb{R}[/tex]
Portanto:
[tex]\Delta >0\Rightarrow \exists x_1, x_2, \ x_1\neq x_2:ax^2+bx+c=0,\ x=x_1 \ ou \ x=x_2[/tex]