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

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

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

We add all the numbers together, and all the variables

(i+3)(-1i+5)=0

We multiply parentheses ..

(-1i^2+5i-3i+15)=0

We get rid of parentheses

-1i^2+5i-3i+15=0

We add all the numbers together, and all the variables

-1i^2+2i+15=0

a = -1; b = 2; c = +15;
Δ = b2-4ac
Δ = 22-4·(-1)·15
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*-1}=\frac{-10}{-2}
=+5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*-1}=\frac{6}{-2}
=-3
$