(2-3i)(4-6i)=1

Solution for (2-3i)(4-6i)=1 equation:

(2-3i)(4-6i)=1

We move all terms to the left:

(2-3i)(4-6i)-(1)=0

We add all the numbers together, and all the variables

(-3i+2)(-6i+4)-1=0

We multiply parentheses ..

(+18i^2-12i-12i+8)-1=0

We get rid of parentheses

18i^2-12i-12i+8-1=0

We add all the numbers together, and all the variables

18i^2-24i+7=0

a = 18; b = -24; c = +7;
Δ = b2-4ac
Δ = -242-4·18·7
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6\sqrt{2}}{2*18}=\frac{24-6\sqrt{2}}{36}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6\sqrt{2}}{2*18}=\frac{24+6\sqrt{2}}{36}
$