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

May 16, 2015

Let us try completing the square:

${\left(x + 0.106\right)}^{2} = {x}^{2} + 0.212 x + 0.011236$

So

$0 = {x}^{2} + 0.212 x - 0.0024$

$= \left({x}^{2} + 0.212 x + 0.011236\right) - 0.011236 - 0.0024$

$= {\left(x + 0.106\right)}^{2} - \left(0.011236 + 0.0024\right)$

$= {\left(x + 0.106\right)}^{2} - 0.013636$

Add $0.013636$ to both sides to get

${\left(x + 0.106\right)}^{2} = 0.013636$

So

$x + 0.106 = \pm \sqrt{0.013636}$

Hence

$x = - 0.106 \pm \sqrt{0.013636}$