If $f\left(x\right)=a{x}^2+bx+c,g\left(x\right)=-a{x}^2+bx+c$ where $ac\ne 0$  then $f\left(x\right)g\left(x\right)=0$ has 

Let $D_1$ and $D_2$ be discriminates of $a{x}^2+bx+c=0$

and $-a{x}^2+bx+c=0$ respectively. Then,  

$D_1=b^2-4ac,D_2=b^2+4ac$ 

Now, $ac\ne 0\Rightarrow$either $ac>0$ or $ac<0$ 

If $ac>0,$then $D_2>0$ Therefore, roots of $-a{x}^2+bx+c=0$ are real . 

If $ac>0,$then $D_1>0$  Therefore, roots of  $a{x}^2+bx+c=0$ are real . 

Thus , $f\left(x\right) g\left(x\right)$ has at least two real roots.