Finding the nature of roots is complex

$f(x)=A{ x }^{ 2 }+Bx+C$ and $A\neq B\neq C$

Let $A+B+C=m\quad \Rightarrow \quad m>0$ ....(1)

and $A-B+C=n\quad \Rightarrow \quad n<0$ ....(2)

From (1), we have $A+C=m-B$

Applying AM-GM on $A$ and $C$ , we have

$\frac { A+C }{ 2 } >\sqrt { AC } \quad \Rightarrow \quad \frac { m-B }{ 2 } >\sqrt { AC }$

Therefore ${ \left( \frac { m-B }{ 2 } \right) }^{ 2 }>AC\quad \Rightarrow \quad { m }^{ 2 }+{ B }^{ 2 }-2mB>4AC$

$\Rightarrow \quad { B }^{ 2 }-4AC>2mB-{ m }^{ 2 }\quad \Rightarrow \quad { B }^{ 2 }-4AC>m(2B-m)$

But ${ B }^{ 2 }-4AC$ is the discriminant of $f(x)$

$\Rightarrow \quad D>m(2B-A-B-C)\quad \Rightarrow \quad D>-m(A-B+C)\\ \Rightarrow \quad D>-mn$

From (1) and (2), $m>0\quad and\quad n<0\quad \Rightarrow \quad mn<0\\ \Rightarrow \quad D>0$

Hence the equation has real and distinct roots.

$B>A+C\implies B^2>A^2+C^2+2AC\implies B^2-4AC>(A-C)^2$

Since $A\ne C$ we have Discriminant real.

Alternative : Let $f(x) = Ax^2+Bx+C$

$f(-x)=Ax^2-Bx+C$

We have two sign changes $f(-x)$ , so that it has two real negative roots by DESCARTE'S RULE OF SIGNS and a quadratic has almost two roots so no complex roots exists

F(1)=A+B+C>0 F(-1)=A-B+C<0 As sign has changed from F(-1)to F(1), there is a root in (-1,1).for A>0 there another root which is <-1 . for A < 0 there is a root greater than one.Hence F(x) has two distinct and real roots.