3x^2+12x-216=0

Solution for 3x^2+12x-216=0 equation:

3x^2+12x-216=0

a = 3; b = 12; c = -216;
Δ = b2-4ac
Δ = 122-4·3·(-216)
Δ = 2736
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:

$\sqrt{\Delta}=\sqrt{2736}=\sqrt{144*19}=\sqrt{144}*\sqrt{19}=12\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{19}}{2*3}=\frac{-12-12\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{19}}{2*3}=\frac{-12+12\sqrt{19}}{6}
$