(2-3i)+5i(1-3i)=0

Solution for (2-3i)+5i(1-3i)=0 equation:

(2-3i)+5i(1-3i)=0

We add all the numbers together, and all the variables

(-3i+2)+5i(-3i+1)=0

We multiply parentheses

-15i^2+(-3i+2)+5i=0

We get rid of parentheses

-15i^2-3i+5i+2=0

We add all the numbers together, and all the variables

-15i^2+2i+2=0

a = -15; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-15)·2
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{31}}{2*-15}=\frac{-2-2\sqrt{31}}{-30}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{31}}{2*-15}=\frac{-2+2\sqrt{31}}{-30}
$