(-4+2i)(i)+(9+5i)=(1-2i)

Solution for (-4+2i)(i)+(9+5i)=(1-2i) equation:

(-4+2i)(i)+(9+5i)=(1-2i)

We move all terms to the left:

(-4+2i)(i)+(9+5i)-((1-2i))=0

We add all the numbers together, and all the variables

(2i-4)i+(5i+9)-((-2i+1))=0

We multiply parentheses

2i^2-4i+(5i+9)-((-2i+1))=0

We get rid of parentheses

2i^2-4i+5i-((-2i+1))+9=0


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

(-2i+1)

We get rid of parentheses

-2i+1

Back to the equation:

-(-2i+1)


We add all the numbers together, and all the variables

2i^2+i-(-2i+1)+9=0

We get rid of parentheses

2i^2+i+2i-1+9=0

We add all the numbers together, and all the variables

2i^2+3i+8=0

a = 2; b = 3; c = +8;
Δ = b2-4ac
Δ = 32-4·2·8
Δ = -55
Delta is less than zero, so there is no solution for the equation