3k(k-12)=0

Solution for 3k(k-12)=0 equation:

3k(k-12)=0

We multiply parentheses

3k^2-36k=0

a = 3; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·3·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*3}=\frac{0}{6}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*3}=\frac{72}{6}
=12
$