3(i)(3+i)=i-10

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Solution for 3(i)(3+i)=i-10 equation: 



3(i)(3+i)=i-10

We move all terms to the left:

3(i)(3+i)-(i-10)=0

We add all the numbers together, and all the variables

3i(i+3)-(i-10)=0

We multiply parentheses

3i^2+9i-(i-10)=0

We get rid of parentheses

3i^2+9i-i+10=0

We add all the numbers together, and all the variables

3i^2+8i+10=0

a = 3; b = 8; c = +10;
Δ = b2-4ac
Δ = 82-4·3·10
Δ = -56
Delta is less than zero, so there is no solution for the equation

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