432=3x^2+2x

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

432=3x^2+2x

We move all terms to the left:

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

We get rid of parentheses

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

The end solution:

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