6(2x-4)=8(x+3)(3x-6)

Solution for 6(2x-4)=8(x+3)(3x-6) equation:

6(2x-4)=8(x+3)(3x-6)

We move all terms to the left:

6(2x-4)-(8(x+3)(3x-6))=0

We multiply parentheses

12x-(8(x+3)(3x-6))-24=0

We multiply parentheses ..

-(8(+3x^2-6x+9x-18))+12x-24=0


We calculate terms in parentheses: -(8(+3x^2-6x+9x-18)), so:

8(+3x^2-6x+9x-18)

We multiply parentheses

24x^2-48x+72x-144

We add all the numbers together, and all the variables

24x^2+24x-144

Back to the equation:

-(24x^2+24x-144)


We add all the numbers together, and all the variables

12x-(24x^2+24x-144)-24=0

We get rid of parentheses

-24x^2+12x-24x+144-24=0

We add all the numbers together, and all the variables

-24x^2-12x+120=0

a = -24; b = -12; c = +120;
Δ = b2-4ac
Δ = -122-4·(-24)·120
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-108}{2*-24}=\frac{-96}{-48}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+108}{2*-24}=\frac{120}{-48}
=-2+1/2
$