The nature of roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0,$ If the expression $ax^2+2bx+c$, where $a$ is a non-zero real number, has the same sign as that of $a$ for every real value of $x$, then roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0$ are:
(A) real and equal
(B) real and unequal
(C) non-real having positive real part
(D) non-real having having negative real part
As the expression $ax^2+2bx+c$ has the same sign as that of $a$ for every real value of $x$, so if $a>0,$ then $4b^2-4ac<0$ and if $a<0$, then $4b^2-4ac>0$
To determine the nature of roots of the equation $ax^2+(b-c)x-2b-c-a=0, I found its discriminant
$\Delta =(b-c)^2+4a(2b+c+a)=b^2+c^2-2bc+8ab+4ac+4a^2$
Now I am not able to find the nature of roots of the equation.

Since $ax^2+2bx+c$ has always the same sign as $a$ for any real $x$, it has no real roots, so $4b^2 - 4ac < 0$.
Now try writing
\begin{align}
(b-c)^2 + 4a(2b+c+a) &= (b-c)^2 + 4(2ab+ac+a^2)  \\
&= (b-c)^2 + 4(2ab+b^2+a^2) + 4(ac - b^2) .
\end{align}
