n(n-3)=22.5

Solution for n(n-3)=22.5 equation:

n(n-3)=22.5

We move all terms to the left:

n(n-3)-(22.5)=0

We add all the numbers together, and all the variables

n(n-3)-22.5=0

We multiply parentheses

n^2-3n-22.5=0

a = 1; b = -3; c = -22.5;
Δ = b2-4ac
Δ = -32-4·1·(-22.5)
Δ = 99
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:

$\sqrt{\Delta}=\sqrt{99}=\sqrt{9*11}=\sqrt{9}*\sqrt{11}=3\sqrt{11}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{11}}{2*1}=\frac{3-3\sqrt{11}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{11}}{2*1}=\frac{3+3\sqrt{11}}{2}
$