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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-10i^2+6i-20i+12)=0

We get rid of parentheses

-10i^2+6i-20i+12=0

We add all the numbers together, and all the variables

-10i^2-14i+12=0

a = -10; b = -14; c = +12;
Δ = b2-4ac
Δ = -142-4·(-10)·12
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-26}{2*-10}=\frac{-12}{-20}
=3/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*-10}=\frac{40}{-20}
=-2
$