Answer
$\left\{1-\sqrt{2},1+\sqrt{2}\right\}$
Work Step by Step
In the form $x^2+bx=c,$ the given equation, $
0.1x^2-0.2x-0.1=0
,$ is equivalent to
\begin{align*}\require{cancel}
(10)(0.1x^2-0.2x-0.1)&=(0)(10)
\\
x^2-2x-1&=0
\\
x^2-2x&=1
.\end{align*}
To complete the square of the expression at the left side, add $\left(\dfrac{b}{2}\right)^2$ to both sides of the equation. That is,
\begin{align*}
x^2-2x+\left(\dfrac{-2}{2}\right)^2&=1+\left(\dfrac{-2}{2}\right)^2
\\\\
x^2-2x+\left(-1\right)^2&=1+\left(-1\right)^2
\\
x^2-2x+1&=1+1
\\
x^2-2x+1&=2
.\end{align*}
Using $a^2\pm2ab+b^2=(a\pm b)^2$ or the factoring of perfect square trinomials, the equation above is equivalent to
\begin{align*}
(x-1)^2&=2
.\end{align*}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{align*}
x-1&=\pm\sqrt{2}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
x&=1\pm\sqrt{2}
.\end{align*}
Hence, the solution set of the equation $
0.1x^2-0.2x-0.1=0
$ is $\left\{1-\sqrt{2},1+\sqrt{2}\right\}$.
