\(a\left(a-b+c\right)< 0\Rightarrow a^2-ab+ac< 0\Rightarrow ac< ab-a^2\)
\(\Rightarrow4ac< 4ab-4a^2\)
Xét \(\Delta=b^2-4ac\)
Do
\(4ac< 4ab-4a^2\Rightarrow\Delta>b^2-\left(4ab-4a^2\right)=b^2-4ab+4a^2=\left(b-2a\right)^2\)
Mà \(\left(b-2a\right)^2\ge0\Rightarrow\Delta>0\Rightarrow\) pt luôn có 2 nghiệm phân biệt