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

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

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

We multiply parentheses ..

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

We get rid of parentheses

w^2+4w+w+4=0

We add all the numbers together, and all the variables

w^2+5w+4=0

a = 1; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·1·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*1}=\frac{-8}{2}
=-4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*1}=\frac{-2}{2}
=-1
$