9((x+3)/2)^2+6((x+3)/2)+1=0

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Solution for 9((x+3)/2)^2+6((x+3)/2)+1=0 equation: 



9((x+3)/2)^2+6((x+3)/2)+1=0

We calculate fractions


(9((x+3)*2))/()+(6((x+3)*2)^2)/()+1=0



We calculate terms in parentheses: +(9((x+3)*2))/(), so:

9((x+3)*2))/(

We multiply all the terms by the denominator


9((x+3)*2))

Back to the equation:

+(9((x+3)*2)))





We calculate terms in parentheses: +(6((x+3)*2)^2)/(), so:

6((x+3)*2)^2)/(

We multiply all the terms by the denominator


6((x+3)*2)^2)

Back to the equation:

+(6((x+3)*2)^2))

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