Find all ordered triples Problem

Suppose $351_7=aca_b$ for positives $a,b,$ such that $b-1=c$. Compute the ordered triple $(a,b,c).$

Here is what I tried:
$351_7 = 183 = a*b^2+c*b+a = a(b^2+1)+c*b = a(b^2+1)+(b-1)*b = b^2*a+b^2-b+a \implies b^2*a+b^2-b+a -183 = 0 \implies 1-4(a)(a-183) \geq 0.$
I am not sure if this helps or not, though.

The inequality $1-4a(a-183) \geq 0$ does not help you finding the solution. Based on your computation I suggest the following approach. 
The equations $351_{7}=aca_{b}$ and $b-1=c$ imply that $183=ab^{2}+b\left( b-1\right) +a$. Solving for $a$ we obtain $$a=\frac{183-b^{2}+b}{ b^{2}+1}.\tag{1} $$ Since  $1\leq a\leq b-1$,  we conclude that $b$ must
satisfy the double inequality
\begin{equation*}
1\leq \frac{183-b^{2}+b}{b^{2}+1}\leq b-1;
\end{equation*}
or the following one, after multiplying through by $b^{2}+1$:
\begin{equation*}
b^{2}+1\leq 183-b^{2}+b\leq b^{3}-b^{2}+b-1.\tag{2}
\end{equation*}
The first inequality in $(2) $ is equivalent to $2b^{2}-b-182\leq 0$, which implies
that $b\leq \left( 1+\sqrt{1457}\right) /4\approx 9.793$. The second
inequality is equivalent to $b^{3}\geq 184$, so $b\geq \sqrt[3]{184}\approx 5.688$.
Therefore the possible values of $b$ are $b\in \left\{ 6,7,8,9\right\} $. However none of these values results in $(1) $ being an integer. 
