0=2(x^2+2x-528)

Solution for 0=2(x^2+2x-528) equation:

0=2(x^2+2x-528)

We move all terms to the left:

0-(2(x^2+2x-528))=0

We add all the numbers together, and all the variables

-(2(x^2+2x-528))=0


We calculate terms in parentheses: -(2(x^2+2x-528)), so:

2(x^2+2x-528)

We multiply parentheses

2x^2+4x-1056

Back to the equation:

-(2x^2+4x-1056)


We get rid of parentheses

-2x^2-4x+1056=0

a = -2; b = -4; c = +1056;
Δ = b2-4ac
Δ = -42-4·(-2)·1056
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-92}{2*-2}=\frac{-88}{-4}
=+22
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+92}{2*-2}=\frac{96}{-4}
=-24
$