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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-5i^2-15i+2i+6)=0

We get rid of parentheses

-5i^2-15i+2i+6=0

We add all the numbers together, and all the variables

-5i^2-13i+6=0

a = -5; b = -13; c = +6;
Δ = b2-4ac
Δ = -132-4·(-5)·6
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*-5}=\frac{-4}{-10}
=2/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*-5}=\frac{30}{-10}
=-3
$