(3+2i)(1+4i)=3+12i+2i+8i2.

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Solution for (3+2i)(1+4i)=3+12i+2i+8i2. equation: 



(3+2i)(1+4i)=3+12i+2i+8i^2.

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

-8i^2.+(+8i^2+2i+12i+3)-12i-2i-3=0

We add all the numbers together, and all the variables

-8i^2.+(+8i^2+2i+12i+3)-14i-3=0

We get rid of parentheses

-8i^2.+8i^2+2i+12i-14i+3-3=0

We add all the numbers together, and all the variables

8i^2-8i^2.=0

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