Question

How do you find the solution set for 0 = x²+0.212x-0.0024 ?

Answer 1

Let us try completing the square:

`(x+0.106)^2 = x^2+0.212x+0.011236`

So

`0 = x^2+0.212x-0.0024`

`= (x^2+0.212x+0.011236)-0.011236-0.0024`

`= (x+0.106)^2-(0.011236+0.0024)`

`= (x+0.106)^2-0.013636`

Add `0.013636` to both sides to get

`(x+0.106)^2 = 0.013636`

So

`x+0.106 = +-sqrt(0.013636)`

Hence

`x = -0.106 +-sqrt(0.013636)`