3n^2+16n+12=0

Solution for 3n^2+16n+12=0 equation:

3n^2+16n+12=0

a = 3; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·3·12
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{7}}{2*3}=\frac{-16-4\sqrt{7}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{7}}{2*3}=\frac{-16+4\sqrt{7}}{6}
$