-0.2x^2-0.3x+0.1=0

Solution for -0.2x^2-0.3x+0.1=0 equation:

-0.2x^2-0.3x+0.1=0

a = -0.2; b = -0.3; c = +0.1;
Δ = b2-4ac
Δ = -0.32-4·(-0.2)·0.1
Δ = 0.17
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{-(-0.3)-\sqrt{0.17}}{2*-0.2}=\frac{0.3-\sqrt{0.17}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.3)+\sqrt{0.17}}{2*-0.2}=\frac{0.3+\sqrt{0.17}}{-0.4}
$