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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+9i^2-15i+12i-20)=0

We get rid of parentheses

9i^2-15i+12i-20=0

We add all the numbers together, and all the variables

9i^2-3i-20=0

a = 9; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·9·(-20)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-27}{2*9}=\frac{-24}{18}
=-1+1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+27}{2*9}=\frac{30}{18}
=1+2/3
$