0=2+8i(2-8i)

Solution for 0=2+8i(2-8i) equation:

0=2+8i(2-8i)

We move all terms to the left:

0-(2+8i(2-8i))=0

We add all the numbers together, and all the variables

-(2+8i(-8i+2))+0=0

We add all the numbers together, and all the variables

-(2+8i(-8i+2))=0


We calculate terms in parentheses: -(2+8i(-8i+2)), so:

2+8i(-8i+2)

determiningTheFunctionDomain
8i(-8i+2)+2

We multiply parentheses

-64i^2+16i+2

Back to the equation:

-(-64i^2+16i+2)


We get rid of parentheses

64i^2-16i-2=0

a = 64; b = -16; c = -2;
Δ = b2-4ac
Δ = -162-4·64·(-2)
Δ = 768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{3}}{2*64}=\frac{16-16\sqrt{3}}{128}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{3}}{2*64}=\frac{16+16\sqrt{3}}{128}
$