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:

\frac{1}{t}+\frac{1}{t-6}=\frac{1}{4}

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 4t(t-6)

4\not{t}(t-6)\cdot\frac{1}{\not{t}}+4t\sout{(t-6)}\cdot\frac{1}{\sout{t-6}}=\not{4}t(t-6)\cdot\frac{1}{\not{4}}

Which when expanding gives us t^{2}-14t+24

Factoring gives (t-12)(t-2)=0

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.

\frac{1}{12}+\frac{1}{6}=\frac{1}{4}

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