(.27)^2+2(.27)(q)+q^2=1

Solution for (.27)^2+2(.27)(q)+q^2=1 equation:

(.27)^2+2(.27)(q)+q^2=1

We move all terms to the left:

(.27)^2+2(.27)(q)+q^2-(1)=0

determiningTheFunctionDomain
q^2+2(.27)q-1+(.27)^2=0

We add all the numbers together, and all the variables

q^2+2(0.27)q-1+(0.27)^2=0

We add all the numbers together, and all the variables

q^2+2(0.27)q-0.9271=0

We multiply parentheses

q^2+0.54q-0.9271=0

a = 1; b = 0.54; c = -0.9271;
Δ = b2-4ac
Δ = 0.542-4·1·(-0.9271)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.54)-2}{2*1}=\frac{-2.54}{2}
=-1+0.54/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.54)+2}{2*1}=\frac{1.46}{2}
=1/1.3698630137
$