0=-2i(8-2i)

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

0=-2i(8-2i)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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


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

-2i(-2i+8)

We multiply parentheses

4i^2-16i

Back to the equation:

-(4i^2-16i)


We get rid of parentheses

-4i^2+16i=0

a = -4; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·(-4)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*-4}=\frac{-32}{-8}
=+4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*-4}=\frac{0}{-8}
=0
$