(4b+b)(b-4)=0

Solution for (4b+b)(b-4)=0 equation:

(4b+b)(b-4)=0

We add all the numbers together, and all the variables

(+5b)(b-4)=0

We multiply parentheses ..

(+5b^2-20b)=0

We get rid of parentheses

5b^2-20b=0

a = 5; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·5·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*5}=\frac{0}{10}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*5}=\frac{40}{10}
=4
$