Question 1161281
What area? The bottom area? The cross section?
Is the sheet of metal square? Rectangular in shape?
A good design is like this, {{{drawing(240, 290,-2,22,-2,27,
rectangle(0,0,20,25),
red(line(4,0,4,25)),red(line(16,0,16,25)),
arrow(7,3,0,3),arrow(13,3,20,3),
locate(7,4,"20(inch)"),locate(8,15,20-2x),
arrow(10,15,4,15),arrow(10,15,16,15),
red(arrow(0,12,4,12)),red(arrow(4,12,0,12)),
red(arrow(16,12,20,12)),red(arrow(20,12,16,12)),
locate(1.5,12,red(x)),locate(18.5,12,red(x))
)}}}with the bends marked as red lines, at {{{x}}} inches from the edges.
Bending up the sides so they are perpendicular to the middle section,
you would get a channel to carry the water.
(Considering other bending angles could not be the intention of the problem, as it would complicate it and lead to an infinite number of solutions)
 
The cross-section would look like this: {{{drawing(200,80,2,18,-2,6,
line(4.2,0,15.8,0),line(4,0.2,4,4),line(16,4,16,0.2),
red(circle(4,0,0.1)),red(circle(16,0,0.1)),
red(circle(4,0.05,0.2)),red(circle(16,0.05,0.2)),
locate(8,0,20-2x),locate(16.3,2.8,red(x))
)}}}
The area of the rectangular cross-section is a function of {{{x}}} ,
{{{A(x)=x(20-2x)}}} {{{inch^2}}} , or {{{A(x)=-2x^2-20x}}} {{{inch^2}}} .
The greater that area, the more water that rain spout can carry per minute.

For a cross-section area of 50 square inches, {{{x}}} should satisfy
{{{-2x^2-20x=50}}} , and that is the quadratic equation to solve.
(The hint about "two answers" suggested a quadratic equation).
 
If you have been shown a way to solve a problem like that in class,
just do it as your teacher suggests.
If that involves transforming that equation into a preferred form,
and/or applying a memorized formula, pleasing the teacher is advisable.
Solving it by factoring or by "completing the square' may be also accepted, and it is easy enough in this case.
 
If you are left to your own resources, you could try different values of {{{x}}} aiming to get {{{A(x)=50}}} .
Those values must be between {{{0}}} and {{{10}}} , because the lengths {{{x}}} and {{{20-2x}}} must be positive).

The function {{{A(x)}}} is a quadratic function with a nice symmetrical graph called a parabola.
The function has a maximum (represented by the top of that symmetrical graph.
The value {{{A=50}}} could be reached at the maximum (one solution),
or at two points to either side of that maximum (2 solutions),
or never (if the maximum value is less than 50).
You realize that {{{A(x)=2x(10-x)}}},
so its value is zero at {{{x=0}}} and {{{x=10}}}
(obvious if you thinks what that means for the cross-section drawing).
Halfway between those two symmetrical points on the graph is the maximum,
with {{{x=(0+10)/2-5}}} and {{{A(x)=2*5(10-5)=2*5*5=50}}} .
So, the get a cross-section area of {{{50}}}{{{inch^2}}} ,
you need to bend up {{{highlight(5inches)}}} on both sides,
and {{{50}}}{{{inch^2}}} is the maximum cross section area possible.