9(1-r)(1+r)=5

Solution for 9(1-r)(1+r)=5 equation:

9(1-r)(1+r)=5

We move all terms to the left:

9(1-r)(1+r)-(5)=0

We add all the numbers together, and all the variables

9(-1r+1)(r+1)-5=0

We multiply parentheses ..

9(-1r^2-1r+r+1)-5=0

We multiply parentheses

-9r^2-9r+9r+9-5=0

We add all the numbers together, and all the variables

-9r^2+4=0

a = -9; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-9)·4
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-9}=\frac{-12}{-18}
=2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-9}=\frac{12}{-18}
=-2/3
$