27k^2-252k+81=0

Solution for 27k^2-252k+81=0 equation:

27k^2-252k+81=0

a = 27; b = -252; c = +81;
Δ = b2-4ac
Δ = -2522-4·27·81
Δ = 54756
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{54756}=234$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-252)-234}{2*27}=\frac{18}{54}
=1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-252)+234}{2*27}=\frac{486}{54}
=9
$