w2+40w=0

Solution for w2+40w=0 equation:

w2+40w=0

We add all the numbers together, and all the variables

w^2+40w=0

a = 1; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·1·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*1}=\frac{-80}{2}
=-40
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*1}=\frac{0}{2}
=0
$