200=n(3n+1)/2

Solution for 200=n(3n+1)/2 equation:

200=n(3n+1)/2

We move all terms to the left:

200-(n(3n+1)/2)=0

We multiply all the terms by the denominator


-(n(3n+1)+200*2)=0


We calculate terms in parentheses: -(n(3n+1)+200*2), so:

n(3n+1)+200*2

We add all the numbers together, and all the variables

n(3n+1)+400

We multiply parentheses

3n^2+n+400

Back to the equation:

-(3n^2+n+400)


We get rid of parentheses

-3n^2-n-400=0

We add all the numbers together, and all the variables

-3n^2-1n-400=0

a = -3; b = -1; c = -400;
Δ = b2-4ac
Δ = -12-4·(-3)·(-400)
Δ = -4799
Delta is less than zero, so there is no solution for the equation