-1/2(20x-12)=-1/5(-10x+35)

Solution for -1/2(20x-12)=-1/5(-10x+35) equation:

-1/2(20x-12)=-1/5(-10x+35)

We move all terms to the left:

-1/2(20x-12)-(-1/5(-10x+35))=0


Domain of the equation: 2(20x-12)!=0

x∈R



Domain of the equation: 5(-10x+35))!=0

x∈R


We calculate fractions


(-5x(-)/(2(20x-12)*5(-10x+35)))+(-(-2x2)/(2(20x-12)*5(-10x+35)))=0


We calculate terms in parentheses: +(-5x(-)/(2(20x-12)*5(-10x+35))), so:

-5x(-)/(2(20x-12)*5(-10x+35))

We add all the numbers together, and all the variables

-5x0/(2(20x-12)*5(-10x+35))

We multiply all the terms by the denominator


-5x0

We add all the numbers together, and all the variables

-5x

Back to the equation:

+(-5x)



We calculate terms in parentheses: +(-(-2x2)/(2(20x-12)*5(-10x+35))), so:

-(-2x2)/(2(20x-12)*5(-10x+35))

We add all the numbers together, and all the variables

-(-2x^2)/(2(20x-12)*5(-10x+35))

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)=0

We get rid of parentheses

2x^2-5x=0

a = 2; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·2·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*2}=\frac{10}{4}
=2+1/2
$