n(n-3)=990

Solution for n(n-3)=990 equation:

n(n-3)=990

We move all terms to the left:

n(n-3)-(990)=0

We multiply parentheses

n^2-3n-990=0

a = 1; b = -3; c = -990;
Δ = b2-4ac
Δ = -32-4·1·(-990)
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3969}=63$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-63}{2*1}=\frac{-60}{2}
=-30
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+63}{2*1}=\frac{66}{2}
=33
$