(5b+1)(4b+1)=0

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

(5b+1)(4b+1)=0

We multiply parentheses ..

(+20b^2+5b+4b+1)=0

We get rid of parentheses

20b^2+5b+4b+1=0

We add all the numbers together, and all the variables

20b^2+9b+1=0

a = 20; b = 9; c = +1;
Δ = b2-4ac
Δ = 92-4·20·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*20}=\frac{-10}{40}
=-1/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*20}=\frac{-8}{40}
=-1/5
$