If the equation,   x2+bx+45=0,b∈Rhas conjugate complex roots and they satisfyz+1=210 , then:

Let  z=α±iβbe roots of the equation So 2α=−b and α2+β2=45 Now, z+1=210⇒α+12+β2=40  ⇒α+12−α2=−5⇒2α+1=-5⇒2α=-6      Hence b=6⇒b2-b=30