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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3x^2-2x+6x-4)=0

We get rid of parentheses

3x^2-2x+6x-4=0

We add all the numbers together, and all the variables

3x^2+4x-4=0

a = 3; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·3·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*3}=\frac{-12}{6}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*3}=\frac{4}{6}
=2/3
$