-0.9x^2+0.1x=0

Solution for -0.9x^2+0.1x=0 equation:

-0.9x^2+0.1x=0

a = -0.9; b = 0.1; c = 0;
Δ = b2-4ac
Δ = 0.12-4·(-0.9)·0
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{-(0.1)-\sqrt{0.01}}{2*-0.9}=\frac{-0.1-\sqrt{0.01}}{-1.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{0.01}}{2*-0.9}=\frac{-0.1+\sqrt{0.01}}{-1.8}
$