w(w+2)=2w(w-2)

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

w(w+2)=2w(w-2)

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(2w(w-2)), so:

2w(w-2)

We multiply parentheses

2w^2-4w

Back to the equation:

-(2w^2-4w)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1w^2+6w=0

a = -1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-1)·0
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{-(6)-6}{2*-1}=\frac{-12}{-2}
=+6
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-1}=\frac{0}{-2}
=0
$