4n^2-4n-28=0

Solution for 4n^2-4n-28=0 equation:

4n^2-4n-28=0

a = 4; b = -4; c = -28;
Δ = b2-4ac
Δ = -42-4·4·(-28)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{29}}{2*4}=\frac{4-4\sqrt{29}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{29}}{2*4}=\frac{4+4\sqrt{29}}{8}
$