(6-6i)(2+4i)=i

Solution for (6-6i)(2+4i)=i equation:

(6-6i)(2+4i)=i

We move all terms to the left:

(6-6i)(2+4i)-(i)=0

We add all the numbers together, and all the variables

(-6i+6)(4i+2)-i=0

We add all the numbers together, and all the variables

-1i+(-6i+6)(4i+2)=0

We multiply parentheses ..

(-24i^2-12i+24i+12)-1i=0

We get rid of parentheses

-24i^2-12i+24i-1i+12=0

We add all the numbers together, and all the variables

-24i^2+11i+12=0

a = -24; b = 11; c = +12;
Δ = b2-4ac
Δ = 112-4·(-24)·12
Δ = 1273
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{-(11)-\sqrt{1273}}{2*-24}=\frac{-11-\sqrt{1273}}{-48}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{1273}}{2*-24}=\frac{-11+\sqrt{1273}}{-48}
$