-4(2i-5)+3(i-4)=-(3i+2(1-4i))+1

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Solution for -4(2i-5)+3(i-4)=-(3i+2(1-4i))+1 equation: 



-4(2i-5)+3(i-4)=-(3i+2(1-4i))+1

We move all terms to the left:

-4(2i-5)+3(i-4)-(-(3i+2(1-4i))+1)=0

We add all the numbers together, and all the variables

-4(2i-5)+3(i-4)-(-(3i+2(-4i+1))+1)=0

We multiply parentheses

-8i+3i-(-(3i+2(-4i+1))+1)+20-12=0



We calculate terms in parentheses: -(-(3i+2(-4i+1))+1), so:

-(3i+2(-4i+1))+1



We calculate terms in parentheses: -(3i+2(-4i+1)), so:

3i+2(-4i+1)

We multiply parentheses

3i-8i+2

We add all the numbers together, and all the variables

-5i+2

Back to the equation:

-(-5i+2)



We get rid of parentheses

5i-2+1

We add all the numbers together, and all the variables

5i-1

Back to the equation:

-(5i-1)



We add all the numbers together, and all the variables

-5i-(5i-1)+8=0

We get rid of parentheses

-5i-5i+1+8=0

We add all the numbers together, and all the variables

-10i+9=0

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

-10i=-9

i=-9/-10

i=9/10

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