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

Solution for -9n^2+9n+9=0 equation:

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

a = -9; b = 9; c = +9;
Δ = b2-4ac
Δ = 92-4·(-9)·9
Δ = 405
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{405}=\sqrt{81*5}=\sqrt{81}*\sqrt{5}=9\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9\sqrt{5}}{2*-9}=\frac{-9-9\sqrt{5}}{-18}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9\sqrt{5}}{2*-9}=\frac{-9+9\sqrt{5}}{-18}
$