(-3-2x+1)^2+(-1-3x-1)^2=100

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Solution for (-3-2x+1)^2+(-1-3x-1)^2=100 equation: 



(-3-2x+1)^2+(-1-3x-1)^2=100

We move all terms to the left:

(-3-2x+1)^2+(-1-3x-1)^2-(100)=0

We add all the numbers together, and all the variables

(-2x-2)^2+(-3x-2)^2-100=0

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

(-2x-2)^2+(-3x-2)^2=100

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