If the equation,   $x^2+bx+45=0,$$\left(b\in R\right)$has conjugate complex roots and they satisfy$\left|z+1\right|=2\sqrt{10}$ , then:

Let  $z=\alpha \pm i\beta$be roots of the equation 

So $2\alpha =-b$ and ${\alpha }^2+{\beta }^2=45$

 Now, $\left|z+1\right|=2\sqrt{10}\Rightarrow$${\left(\alpha +1\right)}^2+{\beta }^2=40$

  $\Rightarrow {\left(\alpha +1\right)}^2-{\alpha }^2=-5$$\Rightarrow 2\alpha +1=-5$$\Rightarrow 2\alpha =-6$     

 Hence $b=6\Rightarrow$$b^2-b=30$