(2k-1)(3k-8)=0

Solution for (2k-1)(3k-8)=0 equation:

(2k-1)(3k-8)=0

We multiply parentheses ..

(+6k^2-16k-3k+8)=0

We get rid of parentheses

6k^2-16k-3k+8=0

We add all the numbers together, and all the variables

6k^2-19k+8=0

a = 6; b = -19; c = +8;
Δ = b2-4ac
Δ = -192-4·6·8
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-13}{2*6}=\frac{6}{12}
=1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+13}{2*6}=\frac{32}{12}
=2+2/3
$