-3/2(2x-4)=1/3(18-6x)

Solution for -3/2(2x-4)=1/3(18-6x) equation:

-3/2(2x-4)=1/3(18-6x)

We move all terms to the left:

-3/2(2x-4)-(1/3(18-6x))=0


Domain of the equation: 2(2x-4)!=0

x∈R



Domain of the equation: 3(18-6x))!=0

x∈R


We add all the numbers together, and all the variables

-3/2(2x-4)-(1/3(-6x+18))=0

We calculate fractions


(-9x(-)/(2(2x-4)*3(-6x+18)))+(-2x2/(2(2x-4)*3(-6x+18)))=0


We calculate terms in parentheses: +(-9x(-)/(2(2x-4)*3(-6x+18))), so:

-9x(-)/(2(2x-4)*3(-6x+18))

We add all the numbers together, and all the variables

-9x0/(2(2x-4)*3(-6x+18))

We multiply all the terms by the denominator


-9x0

We add all the numbers together, and all the variables

-9x

Back to the equation:

+(-9x)



We calculate terms in parentheses: +(-2x2/(2(2x-4)*3(-6x+18))), so:

-2x2/(2(2x-4)*3(-6x+18))

We multiply all the terms by the denominator


-2x2

We add all the numbers together, and all the variables

-2x^2

Back to the equation:

+(-2x^2)


We get rid of parentheses

-2x^2-9x=0

a = -2; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-2)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-2}=\frac{18}{-4}
=-4+1/2
$