If $f(x)=a{x}^{2}+bx+c$ has no real zeros and $a+b+c< 0$, then

Given: $f(x)=ax^2+bx+c$ has no real zeroes and $a+b+c < 0$

To find : $c=0$ or $c > 0$ or $c < 0$

$c\neq 0$

$\because \$ If $c=0$, then $f(x)$ will have real roots which is not true $\left (\because \ x=-\dfrac {-b\pm \sqrt {b^2-4a(0)}}{2a}\right)$

$\because \ a+b+c < 0$

$\Rightarrow \ f(1) < 0$

$\Rightarrow \ f(1)$ lies below $x-$axis $----(I)$

$\because \ f(x)$ has no real roots.

$\therefore \ f(x)\neq 0$ for any real values of $x$

$\Rightarrow \$ graoh of $f(x)$ does not cut $x-$axis$----(II)$

$\Rightarrow \ f(0)=c < 0$ (From $(I)$ & $(II)$)