(1+2i)(6+3i)=15i

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Solution for (1+2i)(6+3i)=15i equation: 



(1+2i)(6+3i)=15i

We move all terms to the left:

(1+2i)(6+3i)-(15i)=0

We add all the numbers together, and all the variables

(2i+1)(3i+6)-15i=0

We add all the numbers together, and all the variables

-15i+(2i+1)(3i+6)=0

We multiply parentheses ..

(+6i^2+12i+3i+6)-15i=0

We get rid of parentheses

6i^2+12i+3i-15i+6=0

We add all the numbers together, and all the variables

6i^2+6=0

a = 6; b = 0; c = +6;
Δ = b2-4ac
Δ = 02-4·6·6
Δ = -144
Delta is less than zero, so there is no solution for the equation

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