Z=(2-3j)(3+4j)

Solution for Z=(2-3j)(3+4j) equation:

=(2-3Z)(3+4Z)

We move all terms to the left:

-((2-3Z)(3+4Z))=0

We add all the numbers together, and all the variables

-((-3Z+2)(4Z+3))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((-12Z^2-9Z+8Z+6)), so:

(-12Z^2-9Z+8Z+6)

We get rid of parentheses

-12Z^2-9Z+8Z+6

We add all the numbers together, and all the variables

-12Z^2-1Z+6

Back to the equation:

-(-12Z^2-1Z+6)


We get rid of parentheses

12Z^2+1Z-6=0

We add all the numbers together, and all the variables

12Z^2+Z-6=0

a = 12; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·12·(-6)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*12}=\frac{-18}{24}
=-3/4
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*12}=\frac{16}{24}
=2/3
$