(9+5i)*(4-2i)=0

Solution for (9+5i)*(4-2i)=0 equation:

(9+5i)(4-2i)=0

We add all the numbers together, and all the variables

(5i+9)(-2i+4)=0

We multiply parentheses ..

(-10i^2+20i-18i+36)=0

We get rid of parentheses

-10i^2+20i-18i+36=0

We add all the numbers together, and all the variables

-10i^2+2i+36=0

a = -10; b = 2; c = +36;
Δ = b2-4ac
Δ = 22-4·(-10)·36
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-38}{2*-10}=\frac{-40}{-20}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+38}{2*-10}=\frac{36}{-20}
=-1+4/5
$