(z+4)^2=3z+12

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Solution for (z+4)^2=3z+12 equation: 



(z+4)^2=3z+12

We move all terms to the left:

(z+4)^2-(3z+12)=0

We get rid of parentheses

(z+4)^2-3z-12=0

We add all the numbers together, and all the variables

-3z+(z+4)^2-12=0

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

-3z+(z+4)^2=12

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