(2b*(5-b))+(5-b)=0

Solution for (2b*(5-b))+(5-b)=0 equation:

(2b(5-b))+(5-b)=0

We add all the numbers together, and all the variables

(2b(-1b+5))+(-1b+5)=0

We get rid of parentheses

(2b(-1b+5))-1b+5=0


We calculate terms in parentheses: +(2b(-1b+5)), so:

2b(-1b+5)

We multiply parentheses

-2b^2+10b

Back to the equation:

+(-2b^2+10b)


We get rid of parentheses

-2b^2+10b-1b+5=0

We add all the numbers together, and all the variables

-2b^2+9b+5=0

a = -2; b = 9; c = +5;
Δ = b2-4ac
Δ = 92-4·(-2)·5
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*-2}=\frac{-20}{-4}
=+5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*-2}=\frac{2}{-4}
=-1/2
$