(z/2)+4=8-(2/z)

Solution for (z/2)+4=8-(2/z) equation:

(z/2)+4=8-(2/z)

We move all terms to the left:

(z/2)+4-(8-(2/z))=0


Domain of the equation: z))!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

(+z/2)-(8-(+2/z))+4=0

We get rid of parentheses

z/2-(8-(+2/z))+4=0

We calculate fractions


z^2/2z+()/2z+4=0

We multiply all the terms by the denominator


z^2+4*2z+()=0

We add all the numbers together, and all the variables

z^2+4*2z=0

Wy multiply elements

z^2+8z=0

a = 1; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*1}=\frac{-16}{2}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*1}=\frac{0}{2}
=0
$