(3+i)(5-8i)=0

Solution for (3+i)(5-8i)=0 equation:

(3+i)(5-8i)=0

We add all the numbers together, and all the variables

(i+3)(-8i+5)=0

We multiply parentheses ..

(-8i^2+5i-24i+15)=0

We get rid of parentheses

-8i^2+5i-24i+15=0

We add all the numbers together, and all the variables

-8i^2-19i+15=0

a = -8; b = -19; c = +15;
Δ = b2-4ac
Δ = -192-4·(-8)·15
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-29}{2*-8}=\frac{-10}{-16}
=5/8
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+29}{2*-8}=\frac{48}{-16}
=-3
$