n(n-3)/2=n+42

Solution for n(n-3)/2=n+42 equation:

n(n-3)/2=n+42

We move all terms to the left:

n(n-3)/2-(n+42)=0

We get rid of parentheses

n(n-3)/2-n-42=0

We multiply all the terms by the denominator


n(n-3)-n*2-42*2=0

We add all the numbers together, and all the variables

n(n-3)-n*2-84=0

We multiply parentheses

n^2-3n-n*2-84=0

Wy multiply elements

n^2-3n-2n-84=0

We add all the numbers together, and all the variables

n^2-5n-84=0

a = 1; b = -5; c = -84;
Δ = b2-4ac
Δ = -52-4·1·(-84)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-19}{2*1}=\frac{-14}{2}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+19}{2*1}=\frac{24}{2}
=12
$