-0.2x^2+0.2=0

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

-0.2x^2+0.2=0

a = -0.2; b = 0; c = +0.2;
Δ = b2-4ac
Δ = 02-4·(-0.2)·0.2
Δ = 0.16
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{0.16}}{2*-0.2}=\frac{0-\sqrt{0.16}}{-0.4}
=-\frac{\sqrt{}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.16}}{2*-0.2}=\frac{0+\sqrt{0.16}}{-0.4}
=\frac{\sqrt{}}{-0.4}
$