A takes 6 days less than the time taken by B to finish a piece of work.If both A and B together can finish it in 4 days,find the time taken by B to finish the work.

I don't see enough information there to use the quadratic equation. You have to know A, B, and C. If this were in the format of "A is 6 less than B, A plus B is 4, what is B?" then the answer would be a=b-6, b+b-6=4, 2b=10, b=5.
But that makes A=-1 which is invalid as a unit of time. I suppose you could use A=-1, b=5, and C=4 for the quadratic equation... let me see what that returns... Nope, that returns nonsese. -.7 and 5.7 as X1 and X2. Is there any more information with this problem?

Think of how much time each can do in one day.

We have the equation:



Solve for t.

Hmm, 1/t-6 = 1/4 - 1/t, 1 = (1/4 - 1/t)(t-6), 1= (t-6)/4 - (t-6)/t, 4 =(t-6)-4(t-6)/t, 4t= (t2-6t)-4t+24, 4t=t^2+24-10t, 0=t^2-14t+24, 0=(t-2)(t-12), I end up with t = 2 or t = 12?
Since using t = 2 results in a negative time unit, t=12? It works when I check it out. 1/12+1/6=1/4. 1/12+2/12= 1/4, 3/12 = 1/4.

It's been a decade since I did any math at this level, so I'm probably doing it the longest possible way since I had to work it out in my head as I went. I just like doing them as puzzles.

That's correct.

One way to do it is to notice that the LCD is



Which when expanding gives us

Factoring gives

It is clear that t=12 or 2. Since 2 is obviously extraneous, we see 12 is the solution.

A does it in 6 days and B in 12.

Reminds me of all of the fun I had in algebra and geometry. Math puzzles are great.