2(2w-4)2w=28

Solution for 2(2w-4)2w=28 equation:

2(2w-4)2w=28

We move all terms to the left:

2(2w-4)2w-(28)=0

We multiply parentheses

8w^2-16w-28=0

a = 8; b = -16; c = -28;
Δ = b2-4ac
Δ = -162-4·8·(-28)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-24\sqrt{2}}{2*8}=\frac{16-24\sqrt{2}}{16}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+24\sqrt{2}}{2*8}=\frac{16+24\sqrt{2}}{16}
$