0=3(x-2)(x+4)

Solution for 0=3(x-2)(x+4) equation:

0=3(x-2)(x+4)

We move all terms to the left:

0-(3(x-2)(x+4))=0

We add all the numbers together, and all the variables

-(3(x-2)(x+4))=0

We multiply parentheses ..

-(3(+x^2+4x-2x-8))=0


We calculate terms in parentheses: -(3(+x^2+4x-2x-8)), so:

3(+x^2+4x-2x-8)

We multiply parentheses

3x^2+12x-6x-24

We add all the numbers together, and all the variables

3x^2+6x-24

Back to the equation:

-(3x^2+6x-24)


We get rid of parentheses

-3x^2-6x+24=0

a = -3; b = -6; c = +24;
Δ = b2-4ac
Δ = -62-4·(-3)·24
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*-3}=\frac{-12}{-6}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*-3}=\frac{24}{-6}
=-4
$