w^2=w+256

Solution for w^2=w+256 equation:

w^2=w+256

We move all terms to the left:

w^2-(w+256)=0

We get rid of parentheses

w^2-w-256=0

We add all the numbers together, and all the variables

w^2-1w-256=0

a = 1; b = -1; c = -256;
Δ = b2-4ac
Δ = -12-4·1·(-256)
Δ = 1025
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:

$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5\sqrt{41}}{2*1}=\frac{1-5\sqrt{41}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5\sqrt{41}}{2*1}=\frac{1+5\sqrt{41}}{2}
$