4n^2+8=12n

Solution for 4n^2+8=12n equation:

4n^2+8=12n

We move all terms to the left:

4n^2+8-(12n)=0

a = 4; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·4·8
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{-(-12)-4}{2*4}=\frac{8}{8}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*4}=\frac{16}{8}
=2
$