There are several numbers that can be multiplied to get 42. For example 6 and 7 can be multiplied and also 13 and 4 can be multiplied to get 42 as a result. But the question also has an additional part and that part requires the multiplying factors to add up and make 17 as the result. In the case of multiplying 6 and 7 we get 42 but on adding them we get 13. So this cannot be the correct answer. Then if we take 13 and 4, we get 42 by multiplication and 17 by addition. So the correct 2 numbers are 13 and 4.

[latex]x_{1,2} = \frac{ -17 \pm \sqrt{ 17 ^2 - 4 \cdot 1 \cdot 42} }{ 2 \cdot 1 }[/latex] [latex]x_{1,2} = \frac{ -17 \pm \sqrt{ 121 } }{ 2 }[/latex] [latex]x_1 = \frac{ -17~+~\sqrt{ 121 } }{ 2 } = -3[/latex] [latex]x_2 = \frac{ -17~-~\sqrt{ 121 } }{ 2 } = -14[/latex] And so [latex]x^2+17x+42 = (x+3)(x+14) \to x = -3, -14[/latex] But that is what x equals to when you solve it, but by doing it this way we just have to give the opposite sign to the numbers. So the two numbers are 3 and 14. They add to 17 and multiply to 42. Hope that helped :)