3k^2-20k+12=0

Solution for 3k^2-20k+12=0 equation:

3k^2-20k+12=0

a = 3; b = -20; c = +12;
Δ = b2-4ac
Δ = -202-4·3·12
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{-(-20)-16}{2*3}=\frac{4}{6}
=2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*3}=\frac{36}{6}
=6
$