(8-3i)(-2-5i)=0

Solution for (8-3i)(-2-5i)=0 equation:

(8-3i)(-2-5i)=0

We add all the numbers together, and all the variables

(-3i+8)(-5i-2)=0

We multiply parentheses ..

(+15i^2+6i-40i-16)=0

We get rid of parentheses

15i^2+6i-40i-16=0

We add all the numbers together, and all the variables

15i^2-34i-16=0

a = 15; b = -34; c = -16;
Δ = b2-4ac
Δ = -342-4·15·(-16)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-46}{2*15}=\frac{-12}{30}
=-2/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+46}{2*15}=\frac{80}{30}
=2+2/3
$