5-2i/(4-3i)+(2+3i)=0

Solution for 5-2i/(4-3i)+(2+3i)=0 equation:

5-2i/(4-3i)+(2+3i)=0


Domain of the equation: (4-3i)!=0

We move all terms containing i to the left, all other terms to the right

-3i!=-4

i!=-4/-3

i!=1+1/3

i∈R


We add all the numbers together, and all the variables

-2i/(-3i+4)+(3i+2)+5=0

We get rid of parentheses

-2i/(-3i+4)+3i+2+5=0

We multiply all the terms by the denominator


-2i+3i*(-3i+4)+2*(-3i+4)+5*(-3i+4)=0

We multiply parentheses

-9i^2-2i+12i-6i-15i+8+20=0

We add all the numbers together, and all the variables

-9i^2-11i+28=0

a = -9; b = -11; c = +28;
Δ = b2-4ac
Δ = -112-4·(-9)·28
Δ = 1129
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{-(-11)-\sqrt{1129}}{2*-9}=\frac{11-\sqrt{1129}}{-18}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{1129}}{2*-9}=\frac{11+\sqrt{1129}}{-18}
$