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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