3k^2-32k+72=0

Solution for 3k^2-32k+72=0 equation:

3k^2-32k+72=0

a = 3; b = -32; c = +72;
Δ = b2-4ac
Δ = -322-4·3·72
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{10}}{2*3}=\frac{32-4\sqrt{10}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{10}}{2*3}=\frac{32+4\sqrt{10}}{6}
$