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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

(+6x^2-4x)

We get rid of parentheses

6x^2-4x

Back to the equation:

-(6x^2-4x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -6; b = 6; c = +4;
Δ = b2-4ac
Δ = 62-4·(-6)·4
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{33}}{2*-6}=\frac{-6-2\sqrt{33}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{33}}{2*-6}=\frac{-6+2\sqrt{33}}{-12}
$