3/2(2x-4)-5x=7-1/5(5-10x)

Solution for 3/2(2x-4)-5x=7-1/5(5-10x) equation:

3/2(2x-4)-5x=7-1/5(5-10x)

We move all terms to the left:

3/2(2x-4)-5x-(7-1/5(5-10x))=0


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

x∈R



Domain of the equation: 5(5-10x))!=0

x∈R


We add all the numbers together, and all the variables

3/2(2x-4)-5x-(7-1/5(-10x+5))=0

We add all the numbers together, and all the variables

-5x+3/2(2x-4)-(7-1/5(-10x+5))=0

We calculate fractions


-5x+(15x(-)/(2(2x-4)*5(-10x+5)))+(-(-2x2)/(2(2x-4)*5(-10x+5)))=0


We calculate terms in parentheses: +(15x(-)/(2(2x-4)*5(-10x+5))), so:

15x(-)/(2(2x-4)*5(-10x+5))

We add all the numbers together, and all the variables

15x0/(2(2x-4)*5(-10x+5))

We multiply all the terms by the denominator


15x0

We add all the numbers together, and all the variables

15x

Back to the equation:

+(15x)



We calculate terms in parentheses: +(-(-2x2)/(2(2x-4)*5(-10x+5))), so:

-(-2x2)/(2(2x-4)*5(-10x+5))

We add all the numbers together, and all the variables

-(-2x^2)/(2(2x-4)*5(-10x+5))

We multiply all the terms by the denominator


-(-2x^2)

We get rid of parentheses

2x^2

Back to the equation:

+(2x^2)


determiningTheFunctionDomain
2x^2-5x+15x=0

We add all the numbers together, and all the variables

2x^2+10x=0

a = 2; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·2·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*2}=\frac{-20}{4}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*2}=\frac{0}{4}
=0
$