3k(k+4)=0

Solution for 3k(k+4)=0 equation:

3k(k+4)=0

We multiply parentheses

3k^2+12k=0

a = 3; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·3·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*3}=\frac{-24}{6}
=-4
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*3}=\frac{0}{6}
=0
$