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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-5i^2+(-2i+3)-20i=0

We get rid of parentheses

-5i^2-2i-20i+3=0

We add all the numbers together, and all the variables

-5i^2-22i+3=0

a = -5; b = -22; c = +3;
Δ = b2-4ac
Δ = -222-4·(-5)·3
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-4\sqrt{34}}{2*-5}=\frac{22-4\sqrt{34}}{-10}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+4\sqrt{34}}{2*-5}=\frac{22+4\sqrt{34}}{-10}
$