Answer
$x=\left\{ \dfrac{1-\sqrt{13}}{2},\dfrac{1+\sqrt{13}}{2} \right\}$
Work Step by Step
Using the properties of equality, the given equation, $
0.1x^2-0.1x=0.3
,$ is equivalent to
\begin{array}{l}\require{cancel}
10(0.1x^2-0.1x)=(0.3)10
\\\\
1x^2-1x=3
\\\\
x^2-x=3
\\\\
x^2-x-3=0
.\end{array}
The equation above has $a=1$, $b=-1$, and $c=-3$
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, the solutions to the equation above are
\begin{array}{l}\require{cancel}
x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-3)}}{2(1)}
\\\\
x=\dfrac{1\pm\sqrt{1+12}}{2}
\\\\
x=\dfrac{1\pm\sqrt{13}}{2}
.\end{array}
Hence, the solutions are $
x=\left\{ \dfrac{1-\sqrt{13}}{2},\dfrac{1+\sqrt{13}}{2} \right\}
.$