3(x-2)+4(2-3x)=2X(3x+8)

Solution for 3(x-2)+4(2-3x)=2X(3x+8) equation:

3(x-2)+4(2-3x)=2x(3x+8)

We move all terms to the left:

3(x-2)+4(2-3x)-(2x(3x+8))=0

We add all the numbers together, and all the variables

3(x-2)+4(-3x+2)-(2x(3x+8))=0

We multiply parentheses

3x-12x-(2x(3x+8))-6+8=0


We calculate terms in parentheses: -(2x(3x+8)), so:

2x(3x+8)

We multiply parentheses

6x^2+16x

Back to the equation:

-(6x^2+16x)


We add all the numbers together, and all the variables

-9x-(6x^2+16x)+2=0

We get rid of parentheses

-6x^2-9x-16x+2=0

We add all the numbers together, and all the variables

-6x^2-25x+2=0

a = -6; b = -25; c = +2;
Δ = b2-4ac
Δ = -252-4·(-6)·2
Δ = 673
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{673}}{2*-6}=\frac{25-\sqrt{673}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{673}}{2*-6}=\frac{25+\sqrt{673}}{-12}
$