Answer:
The two numbers are [tex]3-\sqrt{14}[/tex] and [tex]3+\sqrt{14}[/tex].
Step-by-step explanation:
I am assuming you meant "numbers", otherwise you question has no answer.
Let us call the two numbers [tex]a[/tex] and [tex]b[/tex], then we have
(1). [tex]ab =-5[/tex] (they multiply to -5)
(2). [tex]a+b =6[/tex] (they add to 6}
We solve for [tex]a[/tex] and [tex]b[/tex], by first solving for [tex]a[/tex] in equation (1):
[tex]a = -\dfrac{5}{b}[/tex]
and substituting this value into equation (2):
[tex]-\dfrac{5}{b}+ b =6[/tex]
multiply both sides by [tex]b[/tex] and rearrange to get:
[tex]-5+b^2 =6b[/tex]
[tex]b^2-6b-5=0[/tex]
Using the quadratic formula we solve for [tex]b[/tex] to get:
[tex]b = 3+ \sqrt{14}[/tex]
[tex]b =3-\sqrt{14}[/tex]
which when put into equation (1) give
[tex]a = 3-\sqrt{14}[/tex]
[tex]a = 3+\sqrt{14}[/tex].
Thus, the two numbers are [tex]3-\sqrt{14}[/tex] and [tex]3+\sqrt{14}[/tex].
