If the quadratic equation $a{x}^2+bx+6=0$  does not have real roots and $b\in R^{+}$ then :

$f\left(x\right)=a{x}^2+bx+6=0$ 

Roots are imaginary$\Rightarrow D<0,$  $f\left(x\right)$ has same sign for all$x\in R$

$\therefore b^2-24a<0,f\left(0\right)=6>0,f\left(-1\right)=a-b+6>0$

$\Rightarrow$   $a>b^2/24,a>b-6$

$\Rightarrow a>\max .\left\{b^2/24,b-6\right\}$