5b(4b+1)=0

Solution for 5b(4b+1)=0 equation:

5b(4b+1)=0

We multiply parentheses

20b^2+5b=0

a = 20; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·20·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*20}=\frac{-10}{40}
=-1/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*20}=\frac{0}{40}
=0
$