(-9+5i)(-3+i)=0

Solution for (-9+5i)(-3+i)=0 equation:

(-9+5i)(-3+i)=0

We add all the numbers together, and all the variables

(5i-9)(i-3)=0

We multiply parentheses ..

(+5i^2-15i-9i+27)=0

We get rid of parentheses

5i^2-15i-9i+27=0

We add all the numbers together, and all the variables

5i^2-24i+27=0

a = 5; b = -24; c = +27;
Δ = b2-4ac
Δ = -242-4·5·27
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6}{2*5}=\frac{18}{10}
=1+4/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6}{2*5}=\frac{30}{10}
=3
$