w^2=0.04

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Solution for w^2=0.04 equation:



w^2=0.04
We move all terms to the left:
w^2-(0.04)=0
We add all the numbers together, and all the variables
w^2-0.04=0
a = 1; b = 0; c = -0.04;
Δ = b2-4ac
Δ = 02-4·1·(-0.04)
Δ = 0.16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.16}}{2*1}=\frac{0-\sqrt{0.16}}{2} =-\frac{\sqrt{}}{2} $
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.16}}{2*1}=\frac{0+\sqrt{0.16}}{2} =\frac{\sqrt{}}{2} $

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