n(n-3)=40

Solution for n(n-3)=40 equation:

n(n-3)=40

We move all terms to the left:

n(n-3)-(40)=0

We multiply parentheses

n^2-3n-40=0

a = 1; b = -3; c = -40;
Δ = b2-4ac
Δ = -32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*1}=\frac{-10}{2}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*1}=\frac{16}{2}
=8
$