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

Solution for (n+9)(n-1)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

n^2-1n+9n-9=0

We add all the numbers together, and all the variables

n^2+8n-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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