3n+n^2=162

Solution for 3n+n^2=162 equation:

3n+n^2=162

We move all terms to the left:

3n+n^2-(162)=0

a = 1; b = 3; c = -162;
Δ = b2-4ac
Δ = 32-4·1·(-162)
Δ = 657
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:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{73}}{2*1}=\frac{-3-3\sqrt{73}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{73}}{2*1}=\frac{-3+3\sqrt{73}}{2}
$