((1+i)(1+i))-2i=0

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Solution for ((1+i)(1+i))-2i=0 equation: 



((1+i)(1+i))-2i=0

We add all the numbers together, and all the variables

((i+1)(i+1))-2i=0

We add all the numbers together, and all the variables

-2i+((i+1)(i+1))=0

We multiply parentheses ..

((+i^2+i+i+1))-2i=0



We calculate terms in parentheses: +((+i^2+i+i+1)), so:

(+i^2+i+i+1)

We get rid of parentheses

i^2+i+i+1

We add all the numbers together, and all the variables

i^2+2i+1

Back to the equation:

+(i^2+2i+1)



We add all the numbers together, and all the variables

-2i+(i^2+2i+1)=0

We get rid of parentheses

i^2-2i+2i+1=0

We add all the numbers together, and all the variables

i^2+1=0

a = 1; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·1·1
Δ = -4
Delta is less than zero, so there is no solution for the equation

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