Z^(3/2)=8i

Solution for Z^(3/2)=8i equation:

^(3/2)=8Z

We move all terms to the left:

^(3/2)-(8Z)=0

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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

Wy multiply elements

-16Z^2+3=0

a = -16; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-16)·3
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-16}=\frac{0-8\sqrt{3}}{-32}
=-\frac{8\sqrt{3}}{-32}
=-\frac{\sqrt{3}}{-4}
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-16}=\frac{0+8\sqrt{3}}{-32}
=\frac{8\sqrt{3}}{-32}
=\frac{\sqrt{3}}{-4}
$