0=3x^2+2x-32

Solution for 0=3x^2+2x-32 equation:

0=3x^2+2x-32

We move all terms to the left:

0-(3x^2+2x-32)=0

We add all the numbers together, and all the variables

-(3x^2+2x-32)=0

We get rid of parentheses

-3x^2-2x+32=0

a = -3; b = -2; c = +32;
Δ = b2-4ac
Δ = -22-4·(-3)·32
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{97}}{2*-3}=\frac{2-2\sqrt{97}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{97}}{2*-3}=\frac{2+2\sqrt{97}}{-6}
$