432=3y^2+2y

Solution for 432=3y^2+2y equation:

432=3y^2+2y

We move all terms to the left:

432-(3y^2+2y)=0

We get rid of parentheses

-3y^2-2y+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:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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