The expression $y=a{x}^2+bx+c$ has always the same sign as, c if

Let $f\left(x\right)=a{x}^2+bx+c.$Then $, f\left(0\right)=c$

Thus, the curve $y=f\left(x\right)$meets $y$$\text{-axis }$at $(0,c)$

If $c > 0,$ then by hypothesis$f( x)>0.$ This means that the curve$y=f\left(x\right)$ does not meet x-axis. 

If $c<0,$ then by hypothesis, $f\left(x\right)<0,$ which means that the curve $y=f\left(x\right)$ is always below x-axis and so it does not intersect with x-axis. 

Thus, in both the cases $y=f\left(x\right)$does not intersect with x-axis i.e. $f\left(x\right)\ne 0$ for any real $x.$

Hence, $f\left(x\right)=0$i.e. $a{x}^2+bx+c=0$ has imaginary roots and so we have $b^2<4ac$