4n^2-48=4n

Solution for 4n^2-48=4n equation:

4n^2-48=4n

We move all terms to the left:

4n^2-48-(4n)=0

a = 4; b = -4; c = -48;
Δ = b2-4ac
Δ = -42-4·4·(-48)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-28}{2*4}=\frac{-24}{8}
=-3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+28}{2*4}=\frac{32}{8}
=4
$