Z=(5+i+1-2i)*i

Solution for Z=(5+i+1-2i)*i equation:

=(5+Z+1-2Z)*Z

We move all terms to the left:

-((5+Z+1-2Z)*Z)=0

We add all the numbers together, and all the variables

-((-1Z+6)*Z)=0


We calculate terms in parentheses: -((-1Z+6)*Z), so:

(-1Z+6)*Z

We multiply parentheses

-1Z^2+6Z

Back to the equation:

-(-1Z^2+6Z)


We get rid of parentheses

1Z^2-6Z=0

We add all the numbers together, and all the variables

Z^2-6Z=0

a = 1; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*1}=\frac{0}{2}
=0
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*1}=\frac{12}{2}
=6
$