(3+2i)(4-5i)=0

Solution for (3+2i)(4-5i)=0 equation:

(3+2i)(4-5i)=0

We add all the numbers together, and all the variables

(2i+3)(-5i+4)=0

We multiply parentheses ..

(-10i^2+8i-15i+12)=0

We get rid of parentheses

-10i^2+8i-15i+12=0

We add all the numbers together, and all the variables

-10i^2-7i+12=0

a = -10; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·(-10)·12
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-23}{2*-10}=\frac{-16}{-20}
=4/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+23}{2*-10}=\frac{30}{-20}
=-1+1/2
$