0=(2x-8)(3x+12)

Solution for 0=(2x-8)(3x+12) equation:

0=(2x-8)(3x+12)

We move all terms to the left:

0-((2x-8)(3x+12))=0

We add all the numbers together, and all the variables

-((2x-8)(3x+12))=0

We multiply parentheses ..

-((+6x^2+24x-24x-96))=0


We calculate terms in parentheses: -((+6x^2+24x-24x-96)), so:

(+6x^2+24x-24x-96)

We get rid of parentheses

6x^2+24x-24x-96

We add all the numbers together, and all the variables

6x^2-96

Back to the equation:

-(6x^2-96)


We get rid of parentheses

-6x^2+96=0

a = -6; b = 0; c = +96;
Δ = b2-4ac
Δ = 02-4·(-6)·96
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*-6}=\frac{-48}{-12}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*-6}=\frac{48}{-12}
=-4
$