0.1x^2+0.5x+0.6=0

Solution for 0.1x^2+0.5x+0.6=0 equation:

0.1x^2+0.5x+0.6=0

a = 0.1; b = 0.5; c = +0.6;
Δ = b2-4ac
Δ = 0.52-4·0.1·0.6
Δ = 0.01
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{-(0.5)-\sqrt{0.01}}{2*0.1}=\frac{-0.5-\sqrt{0.01}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.01}}{2*0.1}=\frac{-0.5+\sqrt{0.01}}{0.2}
$