(3+4i)(3-4i)=9

Solution for (3+4i)(3-4i)=9 equation:

(3+4i)(3-4i)=9

We move all terms to the left:

(3+4i)(3-4i)-(9)=0

We add all the numbers together, and all the variables

(4i+3)(-4i+3)-9=0

We multiply parentheses ..

(-16i^2+12i-12i+9)-9=0

We get rid of parentheses

-16i^2+12i-12i+9-9=0

We add all the numbers together, and all the variables

-16i^2=0

a = -16; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-16)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$i=\frac{-b}{2a}=\frac{0}{-32}=0$