w2+48w=0

Solution for w2+48w=0 equation:

w2+48w=0

We add all the numbers together, and all the variables

w^2+48w=0

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

$\sqrt{\Delta}=\sqrt{2304}=48$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*1}=\frac{-96}{2}
=-48
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*1}=\frac{0}{2}
=0
$