Please explain how -2(b+4)(b+5) equals (b+5-2)?

If -2(b+4)(b+5)=(b+5-2) then -2(b^2+5b+4b+20)=b+3
then -2(b^2+9b+20)=b+3 or,-2b^2-18b-40=b+3 then -2b^2-19b-43=0
then 2b^2+19b+43=0.Now you solve for b. [ANSWER]

-2(b + 4)(b + 5)
= -2(b*b + 4*b + b*5 + 4*5)
= -2(b^2 + 4b + 5b + 20)
= -2*b^2 - 2*9b - 2*20
= -2b^2 - 18b - 40

∴ -2(b + 4)(b + 5) ≠ b + 5 - 2

If -2b^2-18b-40=b+3
then,-2b^2-19b-43=0
b=[19 + & - (361-344)^1/2]/-2=-11.56 & -7.438

-2(b+4)(b+5)
this solution is possible by mistake or refer to higher order derivatives and use L-Hospital Rule

how it can be equal?

if you expand the first equation, yield -2b^2 - 18b -40

Sorry to say, but it doesn't.

-2(b+4)(b+5) expands to:

-2b^2 - 16b - 30

sorry!