4n^2-8=4n

Solution for 4n^2-8=4n equation:

4n^2-8=4n

We move all terms to the left:

4n^2-8-(4n)=0

a = 4; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·4·(-8)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*4}=\frac{-8}{8}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*4}=\frac{16}{8}
=2
$