n(n-3)=27

Solution for n(n-3)=27 equation:

n(n-3)=27

We move all terms to the left:

n(n-3)-(27)=0

We multiply parentheses

n^2-3n-27=0

a = 1; b = -3; c = -27;
Δ = b2-4ac
Δ = -32-4·1·(-27)
Δ = 117
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:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{13}}{2*1}=\frac{3-3\sqrt{13}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{13}}{2*1}=\frac{3+3\sqrt{13}}{2}
$