(w-2)(w+)=0

Solution for (w-2)(w+)=0 equation:

(w-2)(w+)=0

We add all the numbers together, and all the variables

(w-2)(+w)=0

We multiply parentheses ..

(+w^2-2w)=0

We get rid of parentheses

w^2-2w=0

a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*1}=\frac{0}{2}
=0
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2}
=2
$