1=p^2+0.8*p+0.16

Solution for 1=p^2+0.8*p+0.16 equation:

1=p^2+0.8p+0.16

We move all terms to the left:

1-(p^2+0.8p+0.16)=0

We get rid of parentheses

-p^2-0.8p-0.16+1=0

We add all the numbers together, and all the variables

-1p^2-0.8p+0.84=0

a = -1; b = -0.8; c = +0.84;
Δ = b2-4ac
Δ = -0.82-4·(-1)·0.84
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.8)-2}{2*-1}=\frac{-1.2}{-2}
=1/1.66666666667
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.8)+2}{2*-1}=\frac{2.8}{-2}
=-1+0.8/2
$