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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((-12Z^2+3Z-8Z+2))=0


We calculate terms in parentheses: -((-12Z^2+3Z-8Z+2)), so:

(-12Z^2+3Z-8Z+2)

We get rid of parentheses

-12Z^2+3Z-8Z+2

We add all the numbers together, and all the variables

-12Z^2-5Z+2

Back to the equation:

-(-12Z^2-5Z+2)


We get rid of parentheses

12Z^2+5Z-2=0

a = 12; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·12·(-2)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*12}=\frac{-16}{24}
=-2/3
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*12}=\frac{6}{24}
=1/4
$