Let's assume the two numbers are $x$ and $y$. The problem provides two conditions:

1. $x \times y = 42$
2. $x + y = -17$

To solve this, we can rewrite the second equation as:

$y = -17 - x$

Substitute this into the first equation:

$x \times (-17 - x) = 42$

Simplify:

$-x^2 - 17x = 42$

Rearrange the equation:

$x^2 + 17x + 42 = 0$

Now, solve this quadratic equation using the quadratic formula:

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

Where $a = 1$, $b = 17$, and $c = 42$. Substituting these values:

$x = \frac{-17 \pm \sqrt{17^2 - 4(1)(42)}}{2(1)}$ $x = \frac{-17 \pm \sqrt{289 - 168}}{2}$ $x = \frac{-17 \pm \sqrt{121}}{2}$ $x = \frac{-17 \pm 11}{2}$

So, the two possible values for $x$ are:

$x = \frac{-17 + 11}{2} = \frac{-6}{2} = -3$ $x = \frac{-17 - 11}{2} = \frac{-28}{2} = -14$

Therefore, the two numbers are $-3$ and $-14$.

Verification:

$-3 \times -14 = 42 \quad \text{and} \quad -3 + (-14) = -17$

Both conditions are satisfied.

Do you need further details or have any questions?

Here are 5 related questions:

1. What two numbers multiply to 30 and add to -11?
2. Can you solve for two numbers that multiply to 56 and add to -15?
3. How can we apply the quadratic formula to solve more complex problems?
4. What are the steps to factor quadratic equations without using the quadratic formula?
5. How does the discriminant in the quadratic formula affect the number of solutions?

Tip: Always verify your solutions by substituting them back into the original equations.