(3+2i)(5-i)=15-3i+10i-2i^2

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Solution for (3+2i)(5-i)=15-3i+10i-2i^2 equation: 



(3+2i)(5-i)=15-3i+10i-2i^2

We move all terms to the left:

(3+2i)(5-i)-(15-3i+10i-2i^2)=0

We add all the numbers together, and all the variables

-(15-3i+10i-2i^2)+(2i+3)(-1i+5)=0

We get rid of parentheses

2i^2+3i-10i+(2i+3)(-1i+5)-15=0

We multiply parentheses ..

2i^2+(-2i^2+10i-3i+15)+3i-10i-15=0

We add all the numbers together, and all the variables

2i^2+(-2i^2+10i-3i+15)-7i-15=0

We get rid of parentheses

2i^2-2i^2+10i-3i-7i+15-15=0

We add all the numbers together, and all the variables

=0

i=0/1

i=0

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