The difference of two numbers is 4 and the product is 140 find the number and form quadratic equation

Let the two numbers be $x$ and $y$.
The difference of the two numbers is 4:

The product of the two numbers is 140:

From equation (1), express $x$ in terms of $y$:

Substitute this expression for $x$ into equation (2):

Expand the equation:

Rearrange the equation into the standard quadratic form $ay^2 + by + c = 0$:

The quadratic equation is $\boxed{y^2 + 4y - 140 = 0}$.

We can solve $y^2 + 4y - 140 = 0$ using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a=1$, $b=4$, $c=-140$.

This gives two possible values for $y$:

Using $x = y + 4$:
If $y = 10$:

If $y = -14$:

The two numbers are either $\boxed{14 \text{ and } 10}$ or $\boxed{-10 \text{ and } -14}$.