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__Hint:__It is clearly a two variable problem. In such questions, try to write one variable in terms of another, and then substitute value of the variable in the second equation. Then you can solve the equation using the quadratic formula.

Here, we have to determine two numbers whose difference is 4 and the product is given as 192.

Let us suppose the two numbers are x and y.

Now, we have difference of two numbers is 4, so we can write equation as

$x-y=4.............\left( 1 \right)$

And, it is given that product of both numbers is 192, so we can write one more equation as;

$xy=192...........\left( 2 \right)$

Now, we can substitute the value of x from equation (1) to equation (2). To get an equation in ‘y’.

From equation (1) we have

$x=y+4$

Now putting value of ‘x’ in equation (2), we get

$\begin{align}

& \left( y+4 \right)\left( y \right)=192 \\

& {{y}^{2}}+4y=192 \\

& {{y}^{2}}+4y-192=0..................\left( 3 \right) \\

\end{align}$

Now, we have a quadratic equation in equation (3).

If we have any quadratic equation $A{{x}^{2}}+Bx+C=0$, then roots of the equation can be given by formula

$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}.........\left( 4 \right)$

Using equation (4) let us calculate roots of equation (3).

Comparing ${{y}^{2}}+4y-192=0\text{ with }A{{x}^{2}}+Bx+C=0$

We get;

A =1, B = 4, C = -192

Therefore, we can get roots as

$\begin{align}

& y=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times 1\times \left( -192 \right)}}{2} \\

& y=\dfrac{-4\pm \sqrt{16+768}}{2} \\

& y=\dfrac{-4\pm \sqrt{784}}{2} \\

\end{align}$

Now, we have $\sqrt{784}=28,$ Hence, y can be given as;

$\begin{align}

& y=\dfrac{-4\pm 28}{2} \\

& Or \\

& y=\dfrac{-32}{2},\dfrac{24}{2} \\

\end{align}$

Hence, we get

$y=-16,12$

Now, we can calculate x with respect to the calculated values of ‘y’ from equation (1);

__Complete step-by-step solution:__Here, we have to determine two numbers whose difference is 4 and the product is given as 192.

Let us suppose the two numbers are x and y.

Now, we have difference of two numbers is 4, so we can write equation as

$x-y=4.............\left( 1 \right)$

And, it is given that product of both numbers is 192, so we can write one more equation as;

$xy=192...........\left( 2 \right)$

Now, we can substitute the value of x from equation (1) to equation (2). To get an equation in ‘y’.

From equation (1) we have

$x=y+4$

Now putting value of ‘x’ in equation (2), we get

$\begin{align}

& \left( y+4 \right)\left( y \right)=192 \\

& {{y}^{2}}+4y=192 \\

& {{y}^{2}}+4y-192=0..................\left( 3 \right) \\

\end{align}$

Now, we have a quadratic equation in equation (3).

If we have any quadratic equation $A{{x}^{2}}+Bx+C=0$, then roots of the equation can be given by formula

$x=\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}.........\left( 4 \right)$

Using equation (4) let us calculate roots of equation (3).

Comparing ${{y}^{2}}+4y-192=0\text{ with }A{{x}^{2}}+Bx+C=0$

We get;

A =1, B = 4, C = -192

Therefore, we can get roots as

$\begin{align}

& y=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times 1\times \left( -192 \right)}}{2} \\

& y=\dfrac{-4\pm \sqrt{16+768}}{2} \\

& y=\dfrac{-4\pm \sqrt{784}}{2} \\

\end{align}$

Now, we have $\sqrt{784}=28,$ Hence, y can be given as;

$\begin{align}

& y=\dfrac{-4\pm 28}{2} \\

& Or \\

& y=\dfrac{-32}{2},\dfrac{24}{2} \\

\end{align}$

Hence, we get

$y=-16,12$

Now, we can calculate x with respect to the calculated values of ‘y’ from equation (1);

Case 1:

y = -16

We have x – y = 4

x + 16 = 4

x = -12

In this case, the two numbers are (-12, -16).

y = -16

We have x – y = 4

x + 16 = 4

x = -12

In this case, the two numbers are (-12, -16).

Case 2:

y =12

As we have x – y = 4

x – 12 = 4

x = 16

In this case, the two numbers are (16, 12).

y =12

As we have x – y = 4

x – 12 = 4

x = 16

In this case, the two numbers are (16, 12).

**Hence, the answer is two sets of numbers - (16, 12) and (-12, -16).**

__Note:__Using the Quadratic Formula is time consuming and prone to errors. One can go wrong while calculating the value of ‘x’ after determining ‘y’ as done in solution. One may not follow the path of substitution. One can mistakenly put wrong values of A, B and C during calculation of roots of quadratic ${{y}^{2}}+4y-192=0$.

A better way to solve such equations is finding the roots by factorization (split the middle term). For the quadratic equation in this question, we could have written (y+16)(y-12) = 0 giving us the possible values of y as 12 or -16.

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