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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((-9x^2-6x+6x+4))=0


We calculate terms in parentheses: -((-9x^2-6x+6x+4)), so:

(-9x^2-6x+6x+4)

We get rid of parentheses

-9x^2-6x+6x+4

We add all the numbers together, and all the variables

-9x^2+4

Back to the equation:

-(-9x^2+4)


We get rid of parentheses

9x^2-4=0

a = 9; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·9·(-4)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*9}=\frac{-12}{18}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*9}=\frac{12}{18}
=2/3
$