4n^2+48=28n

Solution for 4n^2+48=28n equation:

4n^2+48=28n

We move all terms to the left:

4n^2+48-(28n)=0

a = 4; b = -28; c = +48;
Δ = b2-4ac
Δ = -282-4·4·48
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*4}=\frac{24}{8}
=3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*4}=\frac{32}{8}
=4
$