.72+2(.09q)+q^2=1

Solution for .72+2(.09q)+q^2=1 equation:

.72+2(.09q)+q^2=1

We move all terms to the left:

.72+2(.09q)+q^2-(1)=0

We add all the numbers together, and all the variables

q^2+2(+.09q)+.72-1=0

We add all the numbers together, and all the variables

q^2+2(+.09q)-0.28=0

We multiply parentheses

q^2+2q-0.28=0

a = 1; b = 2; c = -0.28;
Δ = b2-4ac
Δ = 22-4·1·(-0.28)
Δ = 5.12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-\sqrt{5.12}}{2*1}=\frac{-2-\sqrt{5.12}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{5.12}}{2*1}=\frac{-2+\sqrt{5.12}}{2}
$