Limit of this recursive sequence and convergence

$$a_{n+1}=\sqrt{4a_n+3}$$ $a_1=5$
I can solve simpler but I get stuck here because I cant find an upper bound or roots of the quadratic equation $a_{n+1} -a_n= \frac{4a_n+3 - a_n^2}{\sqrt{4a_n +3}+a_n}...$ to find monotony.
I tried this generic aproach but have difficulties Convergence and limit of a recursive sequence

Answer

Prove it using induction. Whether the sequence is increasing or decreasing depends on the value of $a_1$. Observe that

$$a_n \le a_{n+1} \implies 4a_n + 3 \le 4a_{n+1} + 3 \implies a_{n+1} \le a_{n+2}$$ and similarly $$a_n \ge a_{n+1} \implies 4a_n + 3 \ge 4a_{n+1} + 3 \implies a_{n+1} \ge a_{n+2}.$$ Thus if $a_1 \le a_2$ the full sequence is nondecreasing and if $a_1 \ge a_2$ the full sequence is nonincreasing.