(3k-4)(k-2)=0

Solution for (3k-4)(k-2)=0 equation:

(3k-4)(k-2)=0

We multiply parentheses ..

(+3k^2-6k-4k+8)=0

We get rid of parentheses

3k^2-6k-4k+8=0

We add all the numbers together, and all the variables

3k^2-10k+8=0

a = 3; b = -10; c = +8;
Δ = b2-4ac
Δ = -102-4·3·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*3}=\frac{8}{6}
=1+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*3}=\frac{12}{6}
=2
$