8(2k+1)(k-1)=0

Solution for 8(2k+1)(k-1)=0 equation:

8(2k+1)(k-1)=0

We multiply parentheses ..

8(+2k^2-2k+k-1)=0

We multiply parentheses

16k^2-16k+8k-8=0

We add all the numbers together, and all the variables

16k^2-8k-8=0

a = 16; b = -8; c = -8;
Δ = b2-4ac
Δ = -82-4·16·(-8)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-24}{2*16}=\frac{-16}{32}
=-1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+24}{2*16}=\frac{32}{32}
=1
$