(1+n)*(n/2)=435

Solution for (1+n)*(n/2)=435 equation:

(1+n)(n/2)=435

We move all terms to the left:

(1+n)(n/2)-(435)=0

We add all the numbers together, and all the variables

(n+1)(+n/2)-435=0

We multiply parentheses ..

(+n^2+n)-435=0

We get rid of parentheses

n^2+n-435=0

a = 1; b = 1; c = -435;
Δ = b2-4ac
Δ = 12-4·1·(-435)
Δ = 1741
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{-(1)-\sqrt{1741}}{2*1}=\frac{-1-\sqrt{1741}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1741}}{2*1}=\frac{-1+\sqrt{1741}}{2}
$