There are n sweets in a bag, 6 of which are red. The rest of the sweets are blue. Jen removes 1 sweet from the bag. Jen then takes another sweet from the bag. The probability that Hannah takes two red sweets is 1/3. Show that n²-n-90=0.

We know the total number of sweets is n, and that 6 are red. When Jen takes the first sweet, the probability of it being red is [number of red sweets] / [how many there are] = 6/n . For the second sweet, the probability is [how many red sweets are left] / [how many sweets in total are left] = 5/(n-1) . So the probabilty of both those things happening is 6/n x 5/(n-1). When multiplying fractions we multiply the top together -> 6x5=30 and the bottom together ->  n x (n-1) = n² - n, giving us 30/(n² - n). We know from the question that this is equal to 1/3 so we set the two equal to each other. 30/(n² - n) =1/3. Then cross multiply, giving us 90 = n² - n. Finally get it all on one side: n² - n - 90 = 0.