(4x+5)/(2x-10)-(6x^2+75)/(3x^2-75)=0

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Solution for (4x+5)/(2x-10)-(6x^2+75)/(3x^2-75)=0 equation: 



(4x+5)/(2x-10)-(6x^2+75)/(3x^2-75)=0



Domain of the equation: (2x-10)!=0

We move all terms containing x to the left, all other terms to the right

2x!=10

x!=10/2

x!=5

x∈R





Domain of the equation: (3x^2-75)!=0

We move all terms containing x to the left, all other terms to the right

3x^2!=75

x^2!=75/3

x^2!=√25

x!=5

x∈R



We calculate fractions


((4x+5)*(3x^2-75))/((2x-10)*(3x^2-75))+(-(6x^2+75)*(2x-10))/((2x-10)*(3x^2-75))=0



We calculate terms in parentheses: +((4x+5)*(3x^2-75))/((2x-10)*(3x^2-75)), so:

(4x+5)*(3x^2-75))/((2x-10)*(3x^2-75)

We multiply all the terms by the denominator


(4x+5)*(3x^2-75))

Back to the equation:

+((4x+5)*(3x^2-75)))





We calculate terms in parentheses: +(-(6x^2+75)*(2x-10))/((2x-10)*(3x^2-75)), so:

-(6x^2+75)*(2x-10))/((2x-10)*(3x^2-75)

We multiply all the terms by the denominator


-(6x^2+75)*(2x-10))

Back to the equation:

+(-(6x^2+75)*(2x-10)))

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