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}

We can assume that $a=1$ and let $f(x) =x^2+bx+c$. 
If $c< 0$ then the graph of $f$ cuts the $y-$ axis under the $x-$ axis, so it must have real roots. 
If $c=0$ then 0ne root is $0$ and the second is $-b$. 
If $c>0$, since we have $$49f({1\over 7})+cf(-1)=0$$
we have 2 possibilities. 
a) If $f(-1)=f(1/7)=0$ we are done.
b) If $f(-1)>0>f(1/7)$ or $f(-1)<0<f(1/7)$ then we have one real root in $(-1,{1\over 7})$ and we are done.  
