3n2+9=165

Solution for 3n2+9=165 equation:

3n^2+9=165

We move all terms to the left:

3n^2+9-(165)=0

We add all the numbers together, and all the variables

3n^2-156=0

a = 3; b = 0; c = -156;
Δ = b2-4ac
Δ = 02-4·3·(-156)
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{13}}{2*3}=\frac{0-12\sqrt{13}}{6}
=-\frac{12\sqrt{13}}{6}
=-2\sqrt{13}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{13}}{2*3}=\frac{0+12\sqrt{13}}{6}
=\frac{12\sqrt{13}}{6}
=2\sqrt{13}
$