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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-6i^2-8i+9i+12)-1=0

We get rid of parentheses

-6i^2-8i+9i+12-1=0

We add all the numbers together, and all the variables

-6i^2+i+11=0

a = -6; b = 1; c = +11;
Δ = b2-4ac
Δ = 12-4·(-6)·11
Δ = 265
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{-(1)-\sqrt{265}}{2*-6}=\frac{-1-\sqrt{265}}{-12}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{265}}{2*-6}=\frac{-1+\sqrt{265}}{-12}
$