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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-10i^2+16i+15i-24)=0

We get rid of parentheses

-10i^2+16i+15i-24=0

We add all the numbers together, and all the variables

-10i^2+31i-24=0

a = -10; b = 31; c = -24;
Δ = b2-4ac
Δ = 312-4·(-10)·(-24)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-1}{2*-10}=\frac{-32}{-20}
=1+3/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+1}{2*-10}=\frac{-30}{-20}
=1+1/2
$