0.1x^2+0.2x-20=0

Solution for 0.1x^2+0.2x-20=0 equation:

0.1x^2+0.2x-20=0

a = 0.1; b = 0.2; c = -20;
Δ = b2-4ac
Δ = 0.22-4·0.1·(-20)
Δ = 8.04
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{-(0.2)-\sqrt{8.04}}{2*0.1}=\frac{-0.2-\sqrt{8.04}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{8.04}}{2*0.1}=\frac{-0.2+\sqrt{8.04}}{0.2}
$