jmc

Since the quadratic has only one solution, the discriminant must be equal to zero. The discriminant is $b^2-4ac=400-4ac=0$, so $ac=\frac{400}{4}=100$. We need to find $a$ and $c$ given $a+c=29$ and $ac=100$. We could write a quadratic equation and solve, but instead we rely on clever algebraic manipulations: Since $a+c=29$, we have $$(a+c{)}^2=a^2+c^2+2ac=29^2=841.$$We subtract $4ac=400$ from each side to find $$a^2+c^2+2ac-4ac=a^2+c^2-2ac=841-400=441.$$We recognize each side as a square, so we take the square root of both sides: $$\sqrt{a^2+c^2-2ac}=\sqrt{(c-a{)}^2}=c-a=\sqrt{441}=21.$$(Technically we should take the positive and negative square root of both sides, but since $c>a$ we know $c-a>0$.) Thus we have $$\begin{array}{rl}c-a& =21\\ c+a& =29\end{array}$$Summing these equations gives $$\begin{array}{rl}2c& =50\\ \Rightarrow c& =25,\end{array}$$and $a=29-c=4$. Thus our ordered pair $(a,c)$ is $\boxed{(4,25)}$.