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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4w^2+21w+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:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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