Read the problem. Draw a picture.

Identify what we are looking for.

We are looking for the length of the diagonal of the corral.

Name what we are looking for.

The diagonal distance from one corner of the corral to the opposite corner is $3$ yards longer than the width of the corral.
The length of the corral is $\frac{5}{2}$ times the width.

Let $w$ = the width of the corral.
$w+3$ = the length of the diagonal of the corral.
$\frac{5w}{2}$ = the length of the corral.

Each half of the corral is a right triangle. We draw a picture of one of them.

Translate into an equation. We can use the Pythagorean theorem to solve for $w$.

Write the Pythagorean theorem.

$a^2+b^2=c^2$

Solve the equation.

Substitute.

${\left(w+3\right)}^2=w^2+{\left(\frac{5w}{2}\right)}^2$

Expand the squares.

$9+6w+w^2=w^2+\frac{25w^2}{4}$

This is a quadratic equation. Rewrite it in standard form.

$\begin{array}{ccccccc}a{x}^2& +& bx& +& c& =& 0\\ \frac{25w^2}{4}& -& 6w& -& 9& =& 0\end{array}$

Solve the equation using the quadratic formula.

Identify the $a$, $b$, $c$ values.

$\begin{array}{ccccccc}a=\frac{25}{4}& & & b=-6& & & c=-9\end{array}$

Write the quadratic formula.

$w=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Then substitute in the values of $a$, $b$, $c$.

$w=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4·\frac{25}{4}\left(-9\right)}}{2·\frac{25}{4}}$

Simplify.

$w=\frac{6\pm \sqrt{36+225}}{\frac{25}{2}}$

$w=\frac{6\pm \sqrt{36+225}}{\frac{25}{2}}$

Rewrite to show two solutions.

$w=\frac{6}{25}\left(2-\sqrt{29}\right)$
$w=\frac{6}{25}\left(2+\sqrt{29}\right)$

Approximate the answers using a calculator.

$w\approx \left(-0.8\right)$
$w\approx 1.8$

We eliminate the negative solution for the width.

Width $w\approx 1.8$

Diagonal
$\approx w+3$
$\approx \left(1.8\right)+3$
$\approx 4.8$

Check the answer.

Check on your own in the Pythagorean theorem.

Answer the question.

The diagonal distance from one corner of the corral to the opposite corner is approximately $4.8$ yards.