9k^2+48k=0

Solution for 9k^2+48k=0 equation:

9k^2+48k=0

a = 9; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·9·0
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*9}=\frac{-96}{18}
=-5+1/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*9}=\frac{0}{18}
=0
$