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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-24i^2+9i-16i+6)=0

We get rid of parentheses

-24i^2+9i-16i+6=0

We add all the numbers together, and all the variables

-24i^2-7i+6=0

a = -24; b = -7; c = +6;
Δ = b2-4ac
Δ = -72-4·(-24)·6
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-25}{2*-24}=\frac{-18}{-48}
=3/8
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+25}{2*-24}=\frac{32}{-48}
=-2/3
$