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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1i^2+3i-3i+9)=0

We get rid of parentheses

-1i^2+3i-3i+9=0

We add all the numbers together, and all the variables

-1i^2+9=0

a = -1; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-1)·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-1}=\frac{-6}{-2}
=+3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-1}=\frac{6}{-2}
=-3
$