0=3x^2+6x-504

Solution for 0=3x^2+6x-504 equation:

0=3x^2+6x-504

We move all terms to the left:

0-(3x^2+6x-504)=0

We add all the numbers together, and all the variables

-(3x^2+6x-504)=0

We get rid of parentheses

-3x^2-6x+504=0

a = -3; b = -6; c = +504;
Δ = b2-4ac
Δ = -62-4·(-3)·504
Δ = 6084
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6084}=78$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-78}{2*-3}=\frac{-72}{-6}
=+12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+78}{2*-3}=\frac{84}{-6}
=-14
$