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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+20i^2-8i-15i+6)=0

We get rid of parentheses

20i^2-8i-15i+6=0

We add all the numbers together, and all the variables

20i^2-23i+6=0

a = 20; b = -23; c = +6;
Δ = b2-4ac
Δ = -232-4·20·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*20}=\frac{16}{40}
=2/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*20}=\frac{30}{40}
=3/4
$