(6+2i)(6-2i)=i

Solution for (6+2i)(6-2i)=i equation:

(6+2i)(6-2i)=i

We move all terms to the left:

(6+2i)(6-2i)-(i)=0

We add all the numbers together, and all the variables

(2i+6)(-2i+6)-i=0

We add all the numbers together, and all the variables

-1i+(2i+6)(-2i+6)=0

We multiply parentheses ..

(-4i^2+12i-12i+36)-1i=0

We get rid of parentheses

-4i^2+12i-12i-1i+36=0

We add all the numbers together, and all the variables

-4i^2-1i+36=0

a = -4; b = -1; c = +36;
Δ = b2-4ac
Δ = -12-4·(-4)·36
Δ = 577
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{577}}{2*-4}=\frac{1-\sqrt{577}}{-8}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{577}}{2*-4}=\frac{1+\sqrt{577}}{-8}
$