The quadratic equation $ax^2+20x+c=4x$ has exactly one solution. If $a+c=20$, and $a<c$ find the ordered pair $(a,c)$.
Subtracting $4x$ from both sides gives us $ax^2 +16x +c = 0.$ Since there is only one solution, the discriminant of this quadratic must be $0$. This gives us
\[16^2 - 4ac = 0 \implies 256 - 4ac = 0 \implies 64 = ac\]
Given $a+c=20$, we solve to get $a=4$ and $c=16$. Thus $(a,c) = \boxed{(4,16)}$.
To find the ordered pair $(a,c)$, we need to solve the given quadratic equation and use the given information.
First, let's rewrite the quadratic equation: $ax^2 + 20x + c = 4x$ can be simplified to $ax^2 + 16x + c = 0$.
Since the equation has exactly one solution, it means the discriminant ($b^2 - 4ac$) must be equal to zero.
The discriminant is found by substituting the values of $a$, $b$, and $c$ into the formula: $b^2 - 4ac = 16^2 - 4(a)(c) = 256 - 4ac$.
Since the discriminant is zero, we have $256 - 4ac = 0$, which can be further simplified to $ac = 64$.
We are also given that $a + c = 20$, so we can rewrite it as $c = 20 - a$.
Substituting this into the equation $ac = 64$, we have $(20 - a)a = 64$.
Simplifying further, we get $20a - a^2 = 64$.
Rearranging the equation, $a^2 - 20a + 64 = 0$.
This is a quadratic equation that can be factored as $(a-16)(a-4) = 0$.
Setting each factor to zero, we have two possible solutions for $a$: $a-16=0$ or $a-4=0$.
For $a-16=0$, solving for $a$, we get $a = 16$.
For $a-4=0$, solving for $a$, we get $a = 4$.
Since $a<c$ and $c = 20 - a$, we can substitute the values of $a$ into $c$.
For $a = 16$, we have $c = 20 - 16 = 4$. So, one ordered pair is $(a,c) = (16, 4)$.
For $a = 4$, we have $c = 20 - 4 = 16$. So, another ordered pair is $(a,c) = (4, 16)$.
Therefore, the two ordered pairs $(a,c)$ are $(16, 4)$ and $(4, 16)$.