(5b+2)(4b-1)=0

Solution for (5b+2)(4b-1)=0 equation:

(5b+2)(4b-1)=0

We multiply parentheses ..

(+20b^2-5b+8b-2)=0

We get rid of parentheses

20b^2-5b+8b-2=0

We add all the numbers together, and all the variables

20b^2+3b-2=0

a = 20; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·20·(-2)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*20}=\frac{-16}{40}
=-2/5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*20}=\frac{10}{40}
=1/4
$