0.2x^2+0.3x-0.2=0

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

0.2x^2+0.3x-0.2=0

a = 0.2; b = 0.3; c = -0.2;
Δ = b2-4ac
Δ = 0.32-4·0.2·(-0.2)
Δ = 0.25
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{0.25}}{2*0.2}=\frac{-0.3-\sqrt{0.25}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.3)+\sqrt{0.25}}{2*0.2}=\frac{-0.3+\sqrt{0.25}}{0.4}
$