w(2w-4)=32

Solution for w(2w-4)=32 equation:

w(2w-4)=32

We move all terms to the left:

w(2w-4)-(32)=0

We multiply parentheses

2w^2-4w-32=0

a = 2; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·2·(-32)
Δ = 272
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{17}}{2*2}=\frac{4-4\sqrt{17}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{17}}{2*2}=\frac{4+4\sqrt{17}}{4}
$