4n-n^2=-96

Solution for 4n-n^2=-96 equation:

4n-n^2=-96

We move all terms to the left:

4n-n^2-(-96)=0

We add all the numbers together, and all the variables

-1n^2+4n+96=0

a = -1; b = 4; c = +96;
Δ = b2-4ac
Δ = 42-4·(-1)·96
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*-1}=\frac{-24}{-2}
=+12
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*-1}=\frac{16}{-2}
=-8
$