2w^2+6w-216=0

Solution for 2w^2+6w-216=0 equation:

2w^2+6w-216=0

a = 2; b = 6; c = -216;
Δ = b2-4ac
Δ = 62-4·2·(-216)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-42}{2*2}=\frac{-48}{4}
=-12
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+42}{2*2}=\frac{36}{4}
=9
$