0.3x2+0.8x+0.2=0

Solution for 0.3x2+0.8x+0.2=0 equation:

0.3x^2+0.8x+0.2=0

a = 0.3; b = 0.8; c = +0.2;
Δ = b2-4ac
Δ = 0.82-4·0.3·0.2
Δ = 0.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.8)-\sqrt{0.4}}{2*0.3}=\frac{-0.8-\sqrt{0.4}}{0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.8)+\sqrt{0.4}}{2*0.3}=\frac{-0.8+\sqrt{0.4}}{0.6}
$