2k(-2+8/k)=0

Solution for 2k(-2+8/k)=0 equation:

2k(-2+8/k)=0


Domain of the equation: k)!=0

k!=0/1

k!=0

k∈R


We add all the numbers together, and all the variables

2k(8/k-2)=0

We multiply parentheses

16k^2-4k=0

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

$\sqrt{\Delta}=\sqrt{16}=4$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*16}=\frac{0}{32}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*16}=\frac{8}{32}
=1/4
$