A=(9+n)(1-n)

Solution for A=(9+n)(1-n) equation:

=(9+A)(1-A)

We move all terms to the left:

-((9+A)(1-A))=0

We add all the numbers together, and all the variables

-((A+9)(-1A+1))=0

We multiply parentheses ..

-((-1A^2+A-9A+9))=0


We calculate terms in parentheses: -((-1A^2+A-9A+9)), so:

(-1A^2+A-9A+9)

We get rid of parentheses

-1A^2+A-9A+9

We add all the numbers together, and all the variables

-1A^2-8A+9

Back to the equation:

-(-1A^2-8A+9)


We get rid of parentheses

1A^2+8A-9=0

We add all the numbers together, and all the variables

A^2+8A-9=0

a = 1; b = 8; c = -9;
Δ = b2-4ac
Δ = 82-4·1·(-9)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-10}{2*1}=\frac{-18}{2}
=-9
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+10}{2*1}=\frac{2}{2}
=1
$