2i(3-2i)+4-i=-3(3-2i)+2i

Solution for 2i(3-2i)+4-i=-3(3-2i)+2i equation:

2i(3-2i)+4-i=-3(3-2i)+2i

We move all terms to the left:

2i(3-2i)+4-i-(-3(3-2i)+2i)=0

We add all the numbers together, and all the variables

2i(-2i+3)-i-(-3(-2i+3)+2i)+4=0

We add all the numbers together, and all the variables

-1i+2i(-2i+3)-(-3(-2i+3)+2i)+4=0

We multiply parentheses

-4i^2-1i+6i-(-3(-2i+3)+2i)+4=0


We calculate terms in parentheses: -(-3(-2i+3)+2i), so:

-3(-2i+3)+2i

We add all the numbers together, and all the variables

2i-3(-2i+3)

We multiply parentheses

2i+6i-9

We add all the numbers together, and all the variables

8i-9

Back to the equation:

-(8i-9)


We add all the numbers together, and all the variables

-4i^2+5i-(8i-9)+4=0

We get rid of parentheses

-4i^2+5i-8i+9+4=0

We add all the numbers together, and all the variables

-4i^2-3i+13=0

a = -4; b = -3; c = +13;
Δ = b2-4ac
Δ = -32-4·(-4)·13
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{217}}{2*-4}=\frac{3-\sqrt{217}}{-8}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{217}}{2*-4}=\frac{3+\sqrt{217}}{-8}
$