b(1-b)+5(1-b)=0

Solution for b(1-b)+5(1-b)=0 equation:

b(1-b)+5(1-b)=0

We add all the numbers together, and all the variables

b(-1b+1)+5(-1b+1)=0

We multiply parentheses

-1b^2+b-5b+5=0

We add all the numbers together, and all the variables

-1b^2-4b+5=0

a = -1; b = -4; c = +5;
Δ = b2-4ac
Δ = -42-4·(-1)·5
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*-1}=\frac{-2}{-2}
=1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*-1}=\frac{10}{-2}
=-5
$