3n^2+47-232=0

Solution for 3n^2+47-232=0 equation:

3n^2+47-232=0

We add all the numbers together, and all the variables

3n^2-185=0

a = 3; b = 0; c = -185;
Δ = b2-4ac
Δ = 02-4·3·(-185)
Δ = 2220
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:

$\sqrt{\Delta}=\sqrt{2220}=\sqrt{4*555}=\sqrt{4}*\sqrt{555}=2\sqrt{555}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{555}}{2*3}=\frac{0-2\sqrt{555}}{6}
=-\frac{2\sqrt{555}}{6}
=-\frac{\sqrt{555}}{3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{555}}{2*3}=\frac{0+2\sqrt{555}}{6}
=\frac{2\sqrt{555}}{6}
=\frac{\sqrt{555}}{3}
$