<s>[INST] Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in R$ will be real if $a(a+7b+49c)+c(a-b+c)<0$ 
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in \mathbb{R}$ will be real if $$a(a+7b+49c)+c(a-b+c)<0$$

My Attempt:
Given
\begin{align}
  a(a+7b+49c)+c(a-b+c) &< 0 \\
  49a \left( \dfrac {a}{49} + \dfrac {b}{7} + c \right)+c(a-b+c) &< 0 \\
  49a \cdot f\left( \dfrac {1}{7} \right) + f(0)\cdot f(-1) &< 0
\end{align}
 [/INST] $$a(a+7b+49c)+c(a-b+c)\lt 0$$
is equivalent to
$$ac\lt \frac{-a^2-7ab+bc-c^2}{50}$$
So, we get
$$\begin{align}b^2-4ac&\gt b^2-\frac{4}{50}(-a^2-7ab+bc-c^2)\\\\&=\frac{1}{50}(50b^2+4a^2+28ab-4bc+4c^2)\\\\&=\frac{1}{50}\left(4\left(a+\frac{7b}{2}\right)^2+(b-2c)^2\right)\ge 0\end{align}$$
</s>