20=1/2(2x+4)(x+3)

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Solution for 20=1/2(2x+4)(x+3) equation: 



20=1/2(2x+4)(x+3)

We move all terms to the left:

20-(1/2(2x+4)(x+3))=0



Domain of the equation: 2(2x+4)(x+3))!=0

x∈R



We multiply parentheses ..

-(1/2(+2x^2+6x+4x+12))+20=0

We multiply all the terms by the denominator


-(1+20*2(+2x^2+6x+4x+12))=0



We calculate terms in parentheses: -(1+20*2(+2x^2+6x+4x+12)), so:

1+20*2(+2x^2+6x+4x+12)

determiningTheFunctionDomain
20*2(+2x^2+6x+4x+12)+1

Wy multiply elements

40x(++1

We use the square of the difference formula

40x(+1

Back to the equation:

-(40x(+1)



We add all the numbers together, and all the variables

-(40x1=0

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