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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4b^2+21b+5=0

a = 4; b = 21; c = +5;
Δ = b2-4ac
Δ = 212-4·4·5
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*4}=\frac{-40}{8}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*4}=\frac{-2}{8}
=-1/4
$