3n^2-16=0

Solution for 3n^2-16=0 equation:

3n^2-16=0

a = 3; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·3·(-16)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*3}=\frac{0-8\sqrt{3}}{6}
=-\frac{8\sqrt{3}}{6}
=-\frac{4\sqrt{3}}{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*3}=\frac{0+8\sqrt{3}}{6}
=\frac{8\sqrt{3}}{6}
=\frac{4\sqrt{3}}{3}
$