The values of p, so that 6 lies between roots of the equation $x^2+2(p-3)x+9=0$ is ____.

$f\left(x\right)=x^2+2(p-3)x+9$ 
$\Delta >0\Rightarrow {\left[2(p-3)\right]}^2-4\left(1\right)\left(9\right)>0$ 
$\Rightarrow 4(p^2-6p+9)-4\left(9\right)>0$ 
$\Rightarrow p^2-6p>0\Rightarrow p<0\text{ or }p>6\to \left(1\right)$ 
$a.f\left(6\right)<0$
$\Rightarrow \left(1\right).\left[36+12(p-3)+9\right]<0$ 
$\Rightarrow 12p+9<0\Rightarrow p<\frac{-3}{4}\to \left(2\right)$ 
From (1) & (2), $p\in (-\infty ,\frac{-3}{4})$