-0.2x^2+0.6x+0.9=-0.5x

Solution for -0.2x^2+0.6x+0.9=-0.5x equation:

-0.2x^2+0.6x+0.9=-0.5x

We move all terms to the left:

-0.2x^2+0.6x+0.9-(-0.5x)=0

We get rid of parentheses

-0.2x^2+0.6x+0.5x+0.9=0

We add all the numbers together, and all the variables

-0.2x^2+1.1x+0.9=0

a = -0.2; b = 1.1; c = +0.9;
Δ = b2-4ac
Δ = 1.12-4·(-0.2)·0.9
Δ = 1.93
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{-(1.1)-\sqrt{1.93}}{2*-0.2}=\frac{-1.1-\sqrt{1.93}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.1)+\sqrt{1.93}}{2*-0.2}=\frac{-1.1+\sqrt{1.93}}{-0.4}
$