Let's denote the two numbers as $x$ and $y$.

We are given the following conditions:

$x + y = 23$ $x \times y = 120$

To find $x$ and $y$, we can solve this system of equations.

First, express $y$ in terms of $x$ using the first equation:

$y = 23 - x$

Substitute this into the second equation:

$x \times (23 - x) = 120$

This expands to:

$23x - x^2 = 120$

Rearrange the equation into a standard quadratic form:

$x^2 - 23x + 120 = 0$

Now, we solve this quadratic equation using the quadratic formula:

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

For our equation $x^2 - 23x + 120 = 0$, the coefficients are $a = 1$, $b = -23$, and $c = 120$. Plugging these into the quadratic formula:

$x = \frac{-(-23) \pm \sqrt{(-23)^2 - 4(1)(120)}}{2(1)}$

$x = \frac{23 \pm \sqrt{529 - 480}}{2}$

$x = \frac{23 \pm \sqrt{49}}{2}$

$x = \frac{23 \pm 7}{2}$

This gives us two possible values for $x$:

$x = \frac{23 + 7}{2} = 15 \quad \text{or} \quad x = \frac{23 - 7}{2} = 8$

Thus, the two numbers are $15$ and $8$.

Let's verify:

• Their sum: $15 + 8 = 23$
• Their product: $15 \times 8 = 120$

Both conditions are satisfied, so the two numbers are $\boxed{15 \text{ and } 8}$.

Would you like more details on the solution or have any other questions?

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Here are five related questions to expand on this topic:

1. How can you solve a system of linear and quadratic equations?
2. What are the different methods to solve a quadratic equation?
3. Can you always factorize a quadratic equation?
4. What is the significance of the discriminant in the quadratic formula?
5. How do you solve a quadratic equation when the discriminant is negative?

Tip: When solving quadratic equations, always check if the equation can be factored easily before using the quadratic formula. This can save time and simplify the process.