-x^2+x=0.19

Solution for -x^2+x=0.19 equation:

-x^2+x=0.19

We move all terms to the left:

-x^2+x-(0.19)=0

We add all the numbers together, and all the variables

-1x^2+x-0.19=0

a = -1; b = 1; c = -0.19;
Δ = b2-4ac
Δ = 12-4·(-1)·(-0.19)
Δ = 0.24
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)-\sqrt{0.24}}{2*-1}=\frac{-1-\sqrt{0.24}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.24}}{2*-1}=\frac{-1+\sqrt{0.24}}{-2}
$