Answer
$x=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2} \right\}
$
Work Step by Step
Using the properties of equality, the given equation, $
0.2x^2+0.4x-0.3=0
,$ is equivalent to
\begin{array}{l}\require{cancel}
10\left(0.2x^2+0.4x-0.3\right)=(0)10
\\\\
2x^2+4x-3=0
.\end{array}
The equation above has $a=2$, $b=4$, 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{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)}
\\\\
x=\dfrac{-4\pm\sqrt{16+24}}{4}
\\\\
x=\dfrac{-4\pm\sqrt{40}}{4}
\\\\
x=\dfrac{-4\pm\sqrt{4\cdot10}}{4}
\\\\
x=\dfrac{-4\pm\sqrt{(2)^2\cdot10}}{4}
\\\\
x=\dfrac{-4\pm\sqrt{(2)^2\cdot10}}{4}
\\\\
x=\dfrac{-4\pm2\sqrt{10}}{4}
\\\\
x=\dfrac{2(-2\pm\sqrt{10})}{4}
\\\\
x=\dfrac{\cancel{2}(-2\pm\sqrt{10})}{\cancel{2}(2)}
\\\\
x=\dfrac{-2\pm\sqrt{10}}{2}
.\end{array}
Hence, the solutions are $
x=\left\{ \dfrac{-2-\sqrt{10}}{2},\dfrac{-2+\sqrt{10}}{2} \right\}
.$