(9w^2)+(w^2)=(10^(-3))

Solution for (9w^2)+(w^2)=(10^(-3)) equation:

(9w^2)+(w^2)=(10^(-3))

We move all terms to the left:

(9w^2)+(w^2)-((10^(-3)))=0

We add all the numbers together, and all the variables

10w^2-0.001=0

a = 10; b = 0; c = -0.001;
Δ = b2-4ac
Δ = 02-4·10·(-0.001)
Δ = 0.04
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{0.04}}{2*10}=\frac{0-\sqrt{0.04}}{20}
=-\frac{\sqrt{}}{20}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.04}}{2*10}=\frac{0+\sqrt{0.04}}{20}
=\frac{\sqrt{}}{20}
$