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