0=2(x^2+2x-520)

Solution for 0=2(x^2+2x-520) equation:

0=2(x^2+2x-520)

We move all terms to the left:

0-(2(x^2+2x-520))=0

We add all the numbers together, and all the variables

-(2(x^2+2x-520))=0


We calculate terms in parentheses: -(2(x^2+2x-520)), so:

2(x^2+2x-520)

We multiply parentheses

2x^2+4x-1040

Back to the equation:

-(2x^2+4x-1040)


We get rid of parentheses

-2x^2-4x+1040=0

a = -2; b = -4; c = +1040;
Δ = b2-4ac
Δ = -42-4·(-2)·1040
Δ = 8336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8336}=\sqrt{16*521}=\sqrt{16}*\sqrt{521}=4\sqrt{521}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{521}}{2*-2}=\frac{4-4\sqrt{521}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{521}}{2*-2}=\frac{4+4\sqrt{521}}{-4}
$