-0.1x^2+0.6-0.5=0

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

-0.1x^2+0.6-0.5=0

We add all the numbers together, and all the variables

-0.1x^2+0.1=0

a = -0.1; b = 0; c = +0.1;
Δ = b2-4ac
Δ = 02-4·(-0.1)·0.1
Δ = 0.04
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)-\sqrt{0.04}}{2*-0.1}=\frac{0-\sqrt{0.04}}{-0.2}
=-\frac{\sqrt{}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.04}}{2*-0.1}=\frac{0+\sqrt{0.04}}{-0.2}
=\frac{\sqrt{}}{-0.2}
$