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

Solution for (8z^2+4z)/2z=0 equation:

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


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R


We multiply all the terms by the denominator


(8z^2+4z)=0

We get rid of parentheses

8z^2+4z=0

a = 8; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·8·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*8}=\frac{-8}{16}
=-1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*8}=\frac{0}{16}
=0
$