-2(x-2)-4x=3x(x+1)-9x

Solution for -2(x-2)-4x=3x(x+1)-9x equation:

-2(x-2)-4x=3x(x+1)-9x

We move all terms to the left:

-2(x-2)-4x-(3x(x+1)-9x)=0

We add all the numbers together, and all the variables

-4x-2(x-2)-(3x(x+1)-9x)=0

We multiply parentheses

-4x-2x-(3x(x+1)-9x)+4=0


We calculate terms in parentheses: -(3x(x+1)-9x), so:

3x(x+1)-9x

We add all the numbers together, and all the variables

-9x+3x(x+1)

We multiply parentheses

3x^2-9x+3x

We add all the numbers together, and all the variables

3x^2-6x

Back to the equation:

-(3x^2-6x)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+4=0

a = -3; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-3)·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-3}=\frac{0-4\sqrt{3}}{-6}
=-\frac{4\sqrt{3}}{-6}
=-\frac{2\sqrt{3}}{-3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-3}=\frac{0+4\sqrt{3}}{-6}
=\frac{4\sqrt{3}}{-6}
=\frac{2\sqrt{3}}{-3}
$