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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-10i^2-6i+40i+24)=0

We get rid of parentheses

-10i^2-6i+40i+24=0

We add all the numbers together, and all the variables

-10i^2+34i+24=0

a = -10; b = 34; c = +24;
Δ = b2-4ac
Δ = 342-4·(-10)·24
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*-10}=\frac{-80}{-20}
=+4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+46}{2*-10}=\frac{12}{-20}
=-3/5
$