3n(4n+12)-2=0

Solution for 3n(4n+12)-2=0 equation:

3n(4n+12)-2=0

We multiply parentheses

12n^2+36n-2=0

a = 12; b = 36; c = -2;
Δ = b2-4ac
Δ = 362-4·12·(-2)
Δ = 1392
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:

$\sqrt{\Delta}=\sqrt{1392}=\sqrt{16*87}=\sqrt{16}*\sqrt{87}=4\sqrt{87}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{87}}{2*12}=\frac{-36-4\sqrt{87}}{24}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{87}}{2*12}=\frac{-36+4\sqrt{87}}{24}
$