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

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

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

We multiply parentheses ..

(+w^2-5w+w-5)=0

We get rid of parentheses

w^2-5w+w-5=0

We add all the numbers together, and all the variables

w^2-4w-5=0

a = 1; b = -4; c = -5;
Δ = b2-4ac
Δ = -42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6}{2*1}=\frac{-2}{2}
=-1
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6}{2*1}=\frac{10}{2}
=5
$