n(n-3)=20

Solution for n(n-3)=20 equation:

n(n-3)=20

We move all terms to the left:

n(n-3)-(20)=0

We multiply parentheses

n^2-3n-20=0

a = 1; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·1·(-20)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{89}}{2*1}=\frac{3-\sqrt{89}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*1}=\frac{3+\sqrt{89}}{2}
$