Recurrence Relation with Square Root

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 $n+1$

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Denote $b_n=\sqrt{1+24a_{n}}$. Then you get $$4b_n^2-4=b_{n+1}^2+6b_{n+1}+5,$$ whence $$2b_n=\pm (b_{n+1}+3).$$
Thank You. That indeed solved the problem as now we can see $$b_{n+1} = 2b_{n} - 3$$ and thus we get $$b_{n} = 2^{n+1} + 3$$ and putting value of $b_{n}$ we can get $a_{n}$ –  Mod May 27 '13 at 13:24