Để pt \(ax^2+bx+c=0\) là pt bậc 2 \(\Leftrightarrow a\ne0\)
Xét \(\Delta=b^2-4ac=b^2-4c.\dfrac{-6c-4b}{5}=b^2+\dfrac{24c^2+16bc}{5}\)
\(=b^2+\dfrac{16}{5}bc+\dfrac{24}{5}c^2=\left(b^2+2.\dfrac{8}{5}bc+\dfrac{64}{25}c^2\right)+\dfrac{56}{25}c^2\)
\(=\left(b+\dfrac{8}{5}c\right)^2+\dfrac{56}{25}c^2\ge0;\forall b;c\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b+\dfrac{8}{5}c=0\\c=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}b=0\\c=0\end{matrix}\right.\) thay vào 5a+4b+6c=0
\(\Rightarrow a=0\) (ktm)
\(\Rightarrow\)Dấu "=" ko xảy ra \(\Rightarrow\Delta>0\)
\(\Rightarrow\) Pt luôn có hai nghiệm pb
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