For the quadratic

Vieta’s formulas give $r+s=6$ and $rs=k$. To use the condition $r^2+s^2=20$, rewrite it as

Substitute the known expressions:

So,

and

To rewrite

in the form

complete the square. Start with the quadratic and group the first two terms:

Half of $-6$ is $-3$, and squaring gives $9$. Add and subtract $9$ inside the expression:

Now factor the perfect square trinomial:

So the equivalent expression is

Because $h(1)=30$ and $h(3)=30$, the parabola has equal heights at times 1 second and 3 seconds, so its axis of symmetry is halfway between them: $t=2$. Also, $h(0)=18$ and $h(4)=18$ confirms the same symmetry. Write the function in vertex form as $h(t)=a(t-2{)}^2+k$.

Use the table values:

and

Subtracting gives

Then

So the equation is

Because $h(0)=0$ and $h(6)=0$, the quadratic can be written as $h(t)=at(t-6)$. Using $h(2)=20$ gives

so $a=-\frac{5}{2}$. The zeros are at $t=0$ and $t=6$, so the axis of symmetry is halfway between them, at

Values at times equally spaced from $t=3$ are equal. Since $2$ and $4$ are each $1$ unit from $3$, it must be true that

If the width is $x$, then the two widths use $2x$ feet of fencing, leaving $40-2x$ feet for the length along the third side. So the area is

The condition that the area is at least $192$ square feet gives

Expanding and rearranging:

Divide by $-2$, reversing the inequality:

Factor:

A product of two factors is less than or equal to $0$ between the roots, inclusive, so