b^2+19b+40=4-b

Solution for b^2+19b+40=4-b equation:

b^2+19b+40=4-b

We move all terms to the left:

b^2+19b+40-(4-b)=0

We add all the numbers together, and all the variables

b^2+19b-(-1b+4)+40=0

We get rid of parentheses

b^2+19b+1b-4+40=0

We add all the numbers together, and all the variables

b^2+20b+36=0

a = 1; b = 20; c = +36;
Δ = b2-4ac
Δ = 202-4·1·36
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*1}=\frac{-36}{2}
=-18
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*1}=\frac{-4}{2}
=-2
$