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

We begin with the quadratic equation $$x^{2}+bx+45=0$$, where $$b\in\mathbb R$$. Its two roots are stated to be conjugate complex numbers, so we may write them as

$$z=\alpha+i\beta\quad\text{and}\quad\bar z=\alpha-i\beta,\qquad\beta\neq0.$$

For a quadratic $$ax^{2}+bx+c=0$$ with roots $$r_{1},r_{2}$$, Vieta’s relations give

$$r_{1}+r_{2}=-\dfrac{b}{a},\qquad r_{1}r_{2}=\dfrac{c}{a}.$$

In our case $$a=1,\;b=b,\;c=45$$, so we have

$$r_{1}+r_{2}=z+\bar z=2\alpha=-b\;\; \Longrightarrow\;\; \alpha=-\dfrac{b}{2}. \quad -(1)$$

and

$$r_{1}r_{2}=z\bar z=\alpha^{2}+\beta^{2}=45. \quad -(2)$$

The roots also satisfy the given geometric condition $$|z+1|=2\sqrt{10}.$$ For any complex number $$w=x+iy$$ we have the modulus formula $$|w|=\sqrt{x^{2}+y^{2}}.$$ Hence

$$|z+1|^{2}=((\alpha+1)+i\beta)((\alpha+1)-i\beta)=(\alpha+1)^{2}+\beta^{2}= (2\sqrt{10})^{2}=40. \quad -(3)$$

Now we possess two equations containing $$\alpha^{2}+\beta^{2}$$ and $$(\alpha+1)^{2}+\beta^{2}$$. Subtracting (2) from (3) eliminates $$\beta^{2}$$:

$$(\alpha+1)^{2}+\beta^{2}-(\alpha^{2}+\beta^{2})=40-45.$$ $$\bigl(\alpha^{2}+2\alpha+1\bigr)-\alpha^{2}= -5.$$ $$2\alpha+1=-5.$$ $$2\alpha=-6.$$ $$\alpha=-3. \quad -(4)$$

Using (1) we connect $$\alpha$$ to $$b$$:

$$-\dfrac{b}{2}=-3\;\;\Longrightarrow\;\; b=6. \quad -(5)$$

With the value of $$b$$ obtained, we examine the option statements. Computing $$b^{2}-b$$ gives

$$b^{2}-b=6^{2}-6=36-6=30. \quad -(6)$$

The option matching equation (6) is A. $$b^{2}-b=30.$$

All other options are easily checked to be inconsistent: $$b^{2}+b=42\neq72,\, b^{2}-b\neq42,\, b^{2}+b\neq12.$$

Hence, the correct answer is Option A.