k(k+9)=0

Solution for k(k+9)=0 equation:

k(k+9)=0

We multiply parentheses

k^2+9k=0

a = 1; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·1·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*1}=\frac{-18}{2}
=-9
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*1}=\frac{0}{2}
=0
$