n(n-3)=162

Solution for n(n-3)=162 equation:

n(n-3)=162

We move all terms to the left:

n(n-3)-(162)=0

We multiply parentheses

n^2-3n-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}
$