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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-8i^2-14i+15=0

a = -8; b = -14; c = +15;
Δ = b2-4ac
Δ = -142-4·(-8)·15
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*-8}=\frac{-12}{-16}
=3/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+26}{2*-8}=\frac{40}{-16}
=-2+1/2
$