4(x+3)^2+5=21

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Solution for 4(x+3)^2+5=21 equation: 



4(x+3)^2+5=21

We move all terms to the left:

4(x+3)^2+5-(21)=0

We add all the numbers together, and all the variables

4(x+3)^2-16=0

We move all terms containing x to the left, all other terms to the right

4(x+3)^2=16

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