Find two numbers that have a sum of $-2$and a product of $-48$.

If $a{x}^2+bx+c=0$, with $a\ne 0$, then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of the equation and then making groups and equating them to zero.

Let $x$and $y$be two numbers. Then the system of equations formed as per the information given is:

role="math" localid="1647850417864" $$\begin{gathered} x+y=-2\dots \left(1\right) \\ x\times y=-48\dots \left(2\right) \end{gathered}$$

Write equation (1) in terms of $x$.

role="math" localid="1647850479875" $$\begin{gathered} x+y=-2 \\ x+y-y=-2-y \\ x=-2-y \end{gathered}$$

Now substitute the value of $x$into equation (2).

$$\begin{gathered} x\times y=-48 \\ (-2-y)·y=-48 \\ -2y-y^2=-48\left[\text{\hspace{0.17em}Distribute\hspace{0.17em}\hspace{0.17em}}\right] \end{gathered}$$

Simplify the quadratic equation for $y$. by completing the square.

$-2y+2y-y^2+y^2=-48+2y+y^2\left[\text{\hspace{0.17em}\hspace{0.17em}make\hspace{0.17em}\hspace{0.17em}one\hspace{0.17em}\hspace{0.17em}side\hspace{0.17em}\hspace{0.17em}equal\hspace{0.17em}\hspace{0.17em}to\hspace{0.17em}\hspace{0.17em}zero\hspace{0.17em}\hspace{0.17em}}\right]$

$$\begin{gathered} 0=y^2+2y-48 \\ 0=(y-6)(y+8)\left[\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Factor\hspace{0.17em}\hspace{0.17em}}\right] \end{gathered}$$

Use the zero product rule to further find the values of $y$.

$y-6=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y+8=0$

$$\begin{gathered} y=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y=-8 \\ x=-2-y \\ x=-2-6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-2-(-8) \\ x=-8\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=6 \end{gathered}$$

The set of numbers is: $(-8,6)$or $(6,-8)$.