A group

I see that you used a different method to solve the equation. You factored out the common factor of 20000 and then used the quadratic formula to find the roots. That’s a valid approach.👏

However, I noticed that you made a mistake in the last step. You wrote n=n 1​=12, but n 1​=2+10=12 is not a solution of the equation. If you plug in n=12 into the equation, you get:

120 = 12² - 5*12

120 = 144 - 60

120 ≠ 84

So n=12 is an extraneous root that does not satisfy the original equation. The only solution that works is n=n 2​=2-10=-8. But since n must be a positive integer, we reject this solution as well. Therefore, there is no solution to the equation using your method.

my solution is:
Let x be the original number of members. Then the original amount each member had to pay is 480,000/x. When four members pulled out, the remaining members had to pay 480,000/(x-4). The difference between these two amounts is 20,000. So we can write the equation:

480,000/x - 480,000/(x-4) = 20,000

To solve this equation, we need to multiply both sides by x(x-4) and simplify:

480,000(x-4) - 480,000x = 20,000x(x-4)

Expanding and collecting like terms, we get:

-1,920,000 = -16,000x² + 80,000x

Dividing both sides by -16,000, we get:

120 = x² - 5x

Adding 25 to both sides and completing the square, we get:

145 = (x-2.5)²

Taking the square root of both sides, we get:

x-2.5 = ±√145

Adding 2.5 to both sides, we get:

x = 2.5 ±√145

Since x must be a positive integer, we reject the negative root and get:

x = 2.5 +√145

Rounding to the nearest whole number, we get:

x = 14

Therefore, the original number of members was 14

48000/(x-4) -480000/x=20000
48000x-48000(x-4)=20000x(x-4)
Divide across with 20000
24x-24(x-4)=x(x-4)
X²-4x+96=0
X²+8x-12x+96=0
X(x+8)-12(x+8)=0
(X-12)(x+8)=0
Either x-12=0

Or x+8=0
Since x cannot be -ve then original number of men is
X=12