3w^2+2w-432=0

Solution for 3w^2+2w-432=0 equation:

3w^2+2w-432=0

a = 3; b = 2; c = -432;
Δ = b2-4ac
Δ = 22-4·3·(-432)
Δ = 5188
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5188}=\sqrt{4*1297}=\sqrt{4}*\sqrt{1297}=2\sqrt{1297}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1297}}{2*3}=\frac{-2-2\sqrt{1297}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1297}}{2*3}=\frac{-2+2\sqrt{1297}}{6}
$