0.3x^2+0.1x-0.9=0

Solution for 0.3x^2+0.1x-0.9=0 equation:

0.3x^2+0.1x-0.9=0

a = 0.3; b = 0.1; c = -0.9;
Δ = b2-4ac
Δ = 0.12-4·0.3·(-0.9)
Δ = 1.09
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{-(0.1)-\sqrt{1.09}}{2*0.3}=\frac{-0.1-\sqrt{1.09}}{0.6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{1.09}}{2*0.3}=\frac{-0.1+\sqrt{1.09}}{0.6}
$