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63=1/2(x+3)(2x+10)
We move all terms to the left:
63-(1/2(x+3)(2x+10))=0
Domain of the equation: 2(x+3)(2x+10))!=0We multiply parentheses ..
x∈R
-(1/2(+2x^2+10x+6x+30))+63=0
We multiply all the terms by the denominator
-(1+63*2(+2x^2+10x+6x+30))=0
We calculate terms in parentheses: -(1+63*2(+2x^2+10x+6x+30)), so:We add all the numbers together, and all the variables
1+63*2(+2x^2+10x+6x+30)
determiningTheFunctionDomain 63*2(+2x^2+10x+6x+30)+1
Wy multiply elements
126x(++1
We use the square of the difference formula
126x(+1
Back to the equation:
-(126x(+1)
-(126x1=0
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