125=1/3(2x-5)(2x-5)(3x)

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Solution for 125=1/3(2x-5)(2x-5)(3x) equation: 



125=1/3(2x-5)(2x-5)(3x)

We move all terms to the left:

125-(1/3(2x-5)(2x-5)(3x))=0



Domain of the equation: 3(2x-5)(2x-5)3x)!=0

x∈R



We multiply parentheses ..

-(1/3(+4x^2-10x-10x+25)3x)+125=0

We multiply all the terms by the denominator


-(1+125*3(+4x^2-10x-10x+25)3x)=0



We calculate terms in parentheses: -(1+125*3(+4x^2-10x-10x+25)3x), so:

1+125*3(+4x^2-10x-10x+25)3x

determiningTheFunctionDomain
125*3(+4x^2-10x-10x+25)3x+1

Wy multiply elements

375x(++1

We use the square of the difference formula

375x(+1

Back to the equation:

-(375x(+1)



We add all the numbers together, and all the variables

-(375x1=0

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