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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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