2k(3k-8)=0

Solution for 2k(3k-8)=0 equation:

2k(3k-8)=0

We multiply parentheses

6k^2-16k=0

a = 6; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·6·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*6}=\frac{0}{12}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*6}=\frac{32}{12}
=2+2/3
$