According to me, that is...

What 2 numbers when added equal 10 and when multiplied equal 130?

Think Algebra 2...

Hrmph. Unless this is some kinda trick question, I don't think this has a tidy answer. I might have done the math wrong, though... Observe:

1. So, x+y=10, and xy=130.

2. Then y=130/x... so, x+(130/x)=10.

3. Multiply both sides by x, and x^2+130=10x.

4. Subtract 10x from both sides, and x^2-10x+130=0.

5. Now for the Quadratic Equasion! x=(-b+-sqrt(b^2-4ac))/2a.

So, x=(-(-10)+-sqrt((-10)^2-(4*1*130)))/(2*1)

6. -(-10) = 10, and those 1's are useless, so let's fix that:

x=(10+-sqrt((-10)^2-(4*130)))/2

7. (-10)^2=100. So, x=(10+-sqrt(100-(4*130)))/2

8. 4*130=520. So, x=(10+-sqrt(100-520))/2

9. 100-520=-420. So, x=(10+-sqrt(-420))/2

10. Uh, oh... Square root of a negative number. That means our results are going to be imaginary.

sqrt(-1)=i, so let's pull that out.

x=(10+-(i*sqrt(420)))/2

11. sqrt(420)=sqrt(4)*sqrt(105).

So, x=(10+-(i*sqrt(4)*sqrt(105)))/2

This may seem pointless, but wait for it...

12. sqrt(4)=2. So x=(10+-(2i*sqrt(105)))/2

13. Now, divide both the top and bottom by 2.

So, x= ((10/2)+-((2i*sqrt(105)/2)))/(2/2)

14. (2i*sqrt(105))/2=i*sqrt(105), 10/2=5, and 2/2=1.

So, x=(5+-i*sqrt(105))/1

15. And anything divided by 1 is itself.

So, x=5+-i*sqrt(105)

16. That leaves us with two possible values for x:

5+i*sqrt(105) and 5-i*sqrt(105)

17. Solve for y:

10-x=y. So 10-(5+i*sqrt(105))=y. So, y=5-i*sqrt(105). Or

10-x=y. So 10-(5-i*sqrt(105))=y. So, y=5+i*sqrt(105).

18. And there's our answers. Either

x=5+i*sqrt(105) and y=5-i*sqrt(105)

or

x=5-i*sqrt(105) and y=5+i*sqrt(105)

Geh. Imaginary numbers. Gotta love 'em.

Any questions, class?