Answer
The solutions are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
Work Step by Step
$0.1x^{2}-0.1x=0.3$
Multiply the whole equation by $10$:
$10(0.1x^{2}-0.1x=0.3)$
$x^{2}-x=3$
Take $3$ to the left side:
$x^{2}-x-3=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=1$, $b=-1$ and $c=−3$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-3)}}{2(1)}=\dfrac{1\pm\sqrt{1+12}}{2}=...$
$...=\dfrac{1\pm\sqrt{13}}{2}=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
The solutions are $x=\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
