Well, I was doing a problem on recurrence relation , where there was given a an recurrence relation and we had to find $a_{n}$ or simplify the recurrence.
The recurrence relation was $$\begin{align} a_{n} &= \frac{1}{16}(1 + 4a_{n+1} + \sqrt{1+24a_{n+1}} ) \\ a_{0} &= 1 \end{align}$$
What I tried was I reached the equation $$4a_{n} - a_{n+1} = \frac{1+\sqrt{1+24a_{n+1}}}{4} $$ Using formula for quadratic roots and putting $$4a_{n} - a_{n+1} = x = \frac{1+\sqrt{1+24a_{n+1}}}{4}$$ $$\implies Ax^2 + Bx + C = 0$$ $$ A = 2, B = -1, C = -3a_{n+1}$$ And then putting the value of $x$ I tried to solve it but nothing fruitful was obtained.
Edit: I made a mistake in the recurrence relation and now $n$ has been swapped by $n+1$
Hint:
Denote $b_n=\sqrt{1+24a_{n}}$. Then you get $$ 4b_n^2-4=b_{n+1}^2+6b_{n+1}+5, $$ whence $$ 2b_n=\pm (b_{n+1}+3). $$