0=3x^2+18x-216

Solution for 0=3x^2+18x-216 equation:

0=3x^2+18x-216

We move all terms to the left:

0-(3x^2+18x-216)=0

We add all the numbers together, and all the variables

-(3x^2+18x-216)=0

We get rid of parentheses

-3x^2-18x+216=0

a = -3; b = -18; c = +216;
Δ = b2-4ac
Δ = -182-4·(-3)·216
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-54}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+54}{2*-3}=\frac{72}{-6}
=-12
$