63=1/2(x+3)(2x+10)

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Solution for 63=1/2(x+3)(2x+10) equation: 



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))!=0

x∈R



We multiply parentheses ..

-(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:

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)



We add all the numbers together, and all the variables

-(126x1=0

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