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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6x^2-4x+6x-4))=0


We calculate terms in parentheses: -((+6x^2-4x+6x-4)), so:

(+6x^2-4x+6x-4)

We get rid of parentheses

6x^2-4x+6x-4

We add all the numbers together, and all the variables

6x^2+2x-4

Back to the equation:

-(6x^2+2x-4)


We get rid of parentheses

-6x^2-2x+4=0

a = -6; b = -2; c = +4;
Δ = b2-4ac
Δ = -22-4·(-6)·4
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*-6}=\frac{-8}{-12}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*-6}=\frac{12}{-12}
=-1
$