-4n^2+4n+99=0

Solution for -4n^2+4n+99=0 equation:

-4n^2+4n+99=0

a = -4; b = 4; c = +99;
Δ = b2-4ac
Δ = 42-4·(-4)·99
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-40}{2*-4}=\frac{-44}{-8}
=5+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+40}{2*-4}=\frac{36}{-8}
=-4+1/2
$