4n^2-4n=99

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

4n^2-4n=99

We move all terms to the left:

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