Given probability, solve unknown number of counters

[CWatters]

Not my homework exactly but it's homework like so..

1. Homework Statement
There are some red and white counters in a bag. At the start there are 7 red and the rest white. Alfie takes two counters at random without putting any back. The probability that the first is white and the second red is 21/80.

How many white counters were in the bag at the start?

Homework Equations

P(W&R) = P(W) * P(R)

The Attempt at a Solution

I have the solution but it took well over the 3.5 mins budgeted for each question in the paper. Along the way I had to use the general equation for solving a quadratic. Did I miss obvious factors of Eqn (3)?

Let: T = Total number of counters, W = number of white counters

T = 7 + W ...........(1)
P(W&R) = (W/T) * (7/(T-1)) = 21/80......(2)

Two equations and 2 unknowns so should be solvable..

Sub for T in (2)...

(W/(7+W)) * (7/(7+W-1) = 21/80
Multiply out..
7W / (W² + 13W + 42) = 21/80
7*80W = 21*(W² + 13W + 42)
80W = 3*(W² + 13W + 42)
80W - 3*(W² + 13W + 42) = 0
80W - 3W² - 39W -126 = 0
finally the quadratic..
3W² - 41W + 126 = 0......(3)

Then using -b+/-Sqrt(b² -4ac)/2a

I got answers 14/3 and 9. Answer is 9 because 14/3 isn't a whole number. I checked it's correct by putting W=9 and T=16 back into (2).

 

[PeroK]

I don't immediately see a shortcut.

 

[haruspex]

Not sure that it is any quicker, but having got to (w+7)(w+6)=80w/3 it is clear that 3|w, and then immediately that 9|w, so can rewrite as (9w'+7)(3w'+2)=80w'. Obviously w'=0 is too small and 1 is too much, so...