4x(2-3x)=2x(2+3x)

Solution for 4x(2-3x)=2x(2+3x) equation:

4x(2-3x)=2x(2+3x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(2x(3x+2)), so:

2x(3x+2)

We multiply parentheses

6x^2+4x

Back to the equation:

-(6x^2+4x)


We get rid of parentheses

-12x^2-6x^2+8x-4x=0

We add all the numbers together, and all the variables

-18x^2+4x=0

a = -18; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-18)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-18}=\frac{-8}{-36}
=2/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-18}=\frac{0}{-36}
=0
$