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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1i^2+5i-5i+25)=0

We get rid of parentheses

-1i^2+5i-5i+25=0

We add all the numbers together, and all the variables

-1i^2+25=0

a = -1; b = 0; c = +25;
Δ = b2-4ac
Δ = 02-4·(-1)·25
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10}{2*-1}=\frac{-10}{-2}
=+5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10}{2*-1}=\frac{10}{-2}
=-5
$