w^2+4=525

Solution for w^2+4=525 equation:

w^2+4=525

We move all terms to the left:

w^2+4-(525)=0

We add all the numbers together, and all the variables

w^2-521=0

a = 1; b = 0; c = -521;
Δ = b2-4ac
Δ = 02-4·1·(-521)
Δ = 2084
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:

$\sqrt{\Delta}=\sqrt{2084}=\sqrt{4*521}=\sqrt{4}*\sqrt{521}=2\sqrt{521}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{521}}{2*1}=\frac{0-2\sqrt{521}}{2}
=-\frac{2\sqrt{521}}{2}
=-\sqrt{521}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{521}}{2*1}=\frac{0+2\sqrt{521}}{2}
=\frac{2\sqrt{521}}{2}
=\sqrt{521}
$