Answer
$\left\{2-\sqrt{3},2+\sqrt{3}\right\}$
Work Step by Step
In the form $x^2+bx=c,$ the given equation, $
0.1p^2-0.4p+0.1=0
,$ is equivalent to
\begin{align*}\require{cancel}
10(0.1p^2-0.4p+0.1)&=(0)10
\\
p^2-4p+1&=0
\\
p^2-4p&=-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*}
p^2-4p+\left(\dfrac{-4}{2}\right)^2&=-1+\left(\dfrac{-4}{2}\right)^2
\\\\
p^2-4p+\left(-2\right)^2&=-1+\left(-2\right)^2
\\
p^2-4p+4&=-1+4
\\
p^2-4p+4&=3
.\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*}
(p-2)^2&=3
.\end{align*}
Taking the square root of both sides (Square Root Principle), the equation above is equivalent to
\begin{align*}
p-2&=\pm\sqrt{3}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}
p&=2\pm\sqrt{3}
.\end{align*}
Hence, the solution set of the equation $
0.1p^2-0.4p+0.1=0
$ is $\left\{2-\sqrt{3},2+\sqrt{3}\right\}$.
