p^2+2p(0.64)+(0.64)^2=1

Solution for p^2+2p(0.64)+(0.64)^2=1 equation:

p^2+2p(0.64)+(0.64)^2=1

We move all terms to the left:

p^2+2p(0.64)+(0.64)^2-(1)=0

determiningTheFunctionDomain
p^2+2p(0.64)-1+(0.64)^2=0

We add all the numbers together, and all the variables

p^2+2p(0.64)-0.5904=0

We multiply parentheses

p^2+1.28p-0.5904=0

a = 1; b = 1.28; c = -0.5904;
Δ = b2-4ac
Δ = 1.282-4·1·(-0.5904)
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{-(1.28)-2}{2*1}=\frac{-3.28}{2}
=-1+1/1.5625
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.28)+2}{2*1}=\frac{0.72}{2}
=0.72/2
$