jmc

We look for a binomial $ax+b$ whose square agrees with $49x^2+56x-64$, except possibly at the constant term. First we note that $a$ must be $7$ or $-7$, since the coefficient of $x^2$ in $(ax+b{)}^2$ is $a^2$, and we need this to equal $49$. Since we are given that $a>0$, we reject $-7$ and select $a=7$.

Now we want $49x^2+56x-64$ to have the same coefficient of $x$ as $(7x+b{)}^2$. Since the coefficient of $x$ in $(7x+b{)}^2$ is $14b$, we solve $56=14b$ to obtain $b=4$. Therefore, $49x^2+56x-64$ agrees with $(7x+4{)}^2$, except that the constant term is different. Specifically, $(7x+4{)}^2=49x^2+56x+16$.

Now we can rewrite Po's original equation as follows: $$\begin{array}{rl}49x^2+56x-64& =0\\ 49x^2+56x+16& =80\\ (7x+4{)}^2& =80.\end{array}$$This gives $a+b+c=7+4+80=\boxed{91}.$