70=1/2(2x+4)(x+5)

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Solution for 70=1/2(2x+4)(x+5) equation: 



70=1/2(2x+4)(x+5)

We move all terms to the left:

70-(1/2(2x+4)(x+5))=0



Domain of the equation: 2(2x+4)(x+5))!=0

x∈R



We multiply parentheses ..

-(1/2(+2x^2+10x+4x+20))+70=0

We multiply all the terms by the denominator


-(1+70*2(+2x^2+10x+4x+20))=0



We calculate terms in parentheses: -(1+70*2(+2x^2+10x+4x+20)), so:

1+70*2(+2x^2+10x+4x+20)

determiningTheFunctionDomain
70*2(+2x^2+10x+4x+20)+1

Wy multiply elements

140x(++1

We use the square of the difference formula

140x(+1

Back to the equation:

-(140x(+1)



We add all the numbers together, and all the variables

-(140x1=0

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