-9n^2-3n+8=8

Solution for -9n^2-3n+8=8 equation:

-9n^2-3n+8=8

We move all terms to the left:

-9n^2-3n+8-(8)=0

We add all the numbers together, and all the variables

-9n^2-3n=0

a = -9; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-9)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-9}=\frac{0}{-18}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-9}=\frac{6}{-18}
=-1/3
$