0=3x^2-20x+12

Solution for 0=3x^2-20x+12 equation:

0=3x^2-20x+12

We move all terms to the left:

0-(3x^2-20x+12)=0

We add all the numbers together, and all the variables

-(3x^2-20x+12)=0

We get rid of parentheses

-3x^2+20x-12=0

a = -3; b = 20; c = -12;
Δ = b2-4ac
Δ = 202-4·(-3)·(-12)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*-3}=\frac{-4}{-6}
=2/3
$