4n2+6=12n

Solution for 4n2+6=12n equation:

4n^2+6=12n

We move all terms to the left:

4n^2+6-(12n)=0

a = 4; b = -12; c = +6;
Δ = b2-4ac
Δ = -122-4·4·6
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*4}=\frac{12-4\sqrt{3}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*4}=\frac{12+4\sqrt{3}}{8}
$