$A+B+C=2149$, Find $A$ In the following form of odd numbers 

If the numbers 
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.

Let's write the numbers as $a_k:=2k-1$, starting at $k=1$. Then, $a_1=1$, $a_2=3$, $a_3=5$, and so on. When $A=a_k$ is in row $n$ (the first row is row $0$), then $B=a_{k+n+1}$ and $C=a_{k+n+2}=B+2$. Hence, we have
\begin{align*}
&& 2149 &= a_k + 2a_{2k+n+1} +2 \\
&\Rightarrow& 2147 &= 2k-1 + 2(2(k+n+1)-1) = 6k + 4n + 1
\end{align*}
Note that the $n$-th row (starting at $n=0$) of the triangle starts with $a_{k_n}$, where
$$k_n = 1+\cdots+n = \frac{n(n+1)}2.$$
Therefore, $k=k_n+q$ where $0\le q\le n$. Substituting, we get
\begin{align*}
&& 2147 &= 3n(n+1) + 10q + 4n + 1 \\
&\Rightarrow &  0 &= 3 n^2 + 7 n + 10 q -2146
\end{align*}
Observe that the linear term $10q$ does not affect the position of the zeros of this quadratic polynomial much. Hence, I just plugged in $q=n/2$ and this polynomial has a zero around $25$. So, $n=25$ seems a good guess.
Indeed, plugging in $n=25$ into our earlier equation gives
\begin{align*}
&& 2147 &= 6*k + 4*25 + 1 \\
&\Rightarrow &  2046 &= 6*k \\
&\Rightarrow &  341 &= k
\end{align*}
So we have $A=a_{341}=681$ in row $n=25$ and $B=2*(341+25+1)-1=733$, $C=735$.
