(W-2)^2=2w^2+2w-3

Solution for (W-2)^2=2w^2+2w-3 equation:

(-2)^2=2W^2+2W-3

We move all terms to the left:

(-2)^2-(2W^2+2W-3)=0

We add all the numbers together, and all the variables

-(2W^2+2W-3)+4=0

We get rid of parentheses

-2W^2-2W+3+4=0

We add all the numbers together, and all the variables

-2W^2-2W+7=0

a = -2; b = -2; c = +7;
Δ = b2-4ac
Δ = -22-4·(-2)·7
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{15}}{2*-2}=\frac{2-2\sqrt{15}}{-4}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{15}}{2*-2}=\frac{2+2\sqrt{15}}{-4}
$