3n2=29

Solution for 3n2=29 equation:

3n^2=29

We move all terms to the left:

3n^2-(29)=0

a = 3; b = 0; c = -29;
Δ = b2-4ac
Δ = 02-4·3·(-29)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{87}}{2*3}=\frac{0-2\sqrt{87}}{6}
=-\frac{2\sqrt{87}}{6}
=-\frac{\sqrt{87}}{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{87}}{2*3}=\frac{0+2\sqrt{87}}{6}
=\frac{2\sqrt{87}}{6}
=\frac{\sqrt{87}}{3}
$