4n^2=12n

Solution for 4n^2=12n equation:

4n^2=12n

We move all terms to the left:

4n^2-(12n)=0

a = 4; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·4·0
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{-(-12)-12}{2*4}=\frac{0}{8}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*4}=\frac{24}{8}
=3
$