(-p^2)+p-0.1=0

Solution for (-p^2)+p-0.1=0 equation:

(-p^2)+p-0.1=0

We get rid of parentheses

-p^2+p-0.1=0

We add all the numbers together, and all the variables

-1p^2+p-0.1=0

a = -1; b = 1; c = -0.1;
Δ = b2-4ac
Δ = 12-4·(-1)·(-0.1)
Δ = 0.6
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{0.6}}{2*-1}=\frac{-1-\sqrt{0.6}}{-2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.6}}{2*-1}=\frac{-1+\sqrt{0.6}}{-2}
$