SOLUTION: In a​ round-robin chess​ tournament, each player is paired with every other player once. The​ function, shown​ below, models the number of chess​ game

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Question 1066884: In a​ round-robin chess​ tournament, each player is paired with every other player once. The​ function, shown​ below, models the number of chess​ games, N, that must be played in a​ round-robin tournament with t chess players. In a​ round-robin chess​ tournament, 6 games were played. How many players entered the​tournament?
how to solve formula? N=t^2-t / 2
Found 2 solutions by jim_thompson5910, ikleyn:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
N+=+%28t%5E2-t%29%2F2


6+=+%28t%5E2-t%29%2F2 Replace N with 6. The goal is to solve for t.


2%2A6+=+2%2A%28%28t%5E2-t%29%2F2%29 Multiply both sides by 2


12+=+t%5E2-t


12-12+=+t%5E2-t-12 Subtract 12 from both sides


0+=+t%5E2-t-12


t%5E2-t-12+=+0


%28t-4%29%28t%2B3%29+=+0 Factor


t-4=0 or t%2B3=0


t=4 or t=-3


Ignore the negative solution since it makes no sense to have -3 players.


The only practical solution is t+=+4


There are 4 players in the tournament.

Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.
In a​ round-robin chess​ tournament, each player is paired with every other player once.
The​ function, shown​ below, models the number of chess​ games, N, that must be played in a​
round-robin tournament with t chess players. In a​ round-robin chess​ tournament, 6 games were played.
How many players entered the​tournament?
how to solve formula? N=t^2-t / 2
~~~~~~~~~~~~~~~~~~~~~~ 

You need to solve the equation %28t%5E2-t%29%2F2 = 6.

Multiply both sides by 2. You will get an equivalent equations

t%5E2-t = 12,

t%5E2+-t+-+12 = 0,

(t-4)*(t+3) = 0.

The roots are t= 4 and t= -3.

Since you need integer positive solution, the only one is t= 4.


Answer. 4 players.