#### Answer

The solutions are $x=-1\pm\dfrac{\sqrt{10}}{2}$

#### Work Step by Step

$0.2x^{2}+0.4x-0.3=0$
Multiply the whole equation by $10$:
$10(0.2x^{2}+0.4x-0.3=0)$
$2x^{2}+4x-3=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=2$, $b=4$ and $c=-3$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-4\pm\sqrt{4^{2}-4(2)(-3)}}{2(2)}=\dfrac{-4\pm\sqrt{16+24}}{4}=...$
$...=\dfrac{-4\pm\sqrt{40}}{4}=\dfrac{-4\pm2\sqrt{10}}{4}=-\dfrac{4}{4}\pm\dfrac{2\sqrt{10}}{4}=...$
$...=-1\pm\dfrac{\sqrt{10}}{2}$
The solutions are $x=-1\pm\dfrac{\sqrt{10}}{2}$