Re: Let's talk Quadratic Equations
Yes, the signs do get confusing - there are two ways of solving a problem like this. In your problem, (the 2 in the parentheses means squared, I can't do superscript on vBulletin) 2x(2) + 4x - 30 = 0, you could save yourself some trouble and divide everything by two, but I don't know if this is standard. I'd just divide out what I could.
At any rate, in 2x(2) + 4x - 30 = 0, in ax(2) + bx + c = 0 form, you understand how a and c have to be put together in a way to make the middle term. You know this, so I'll skip the discovering that step.
Now we know that a, which is, in this case 2, is 1 and 2, as that is the only multiplication duo (in whole numbers) that will work. As well, using deduction, we know that c, 30, is 3 and 10. So we have four numbers:
a.....c
1.....3
2.....10
We know that a is either both positive or both negative because it has to equal a positive 2, and we know that in c one has to be positive and the other has to be negative because it has to equal a -30, and so it works out like this:
..a.......c
+1.....-3
+2.....+10
We know we have to make 2 and -3 go together to get -6, and we know that 1 and +10 have to go together to get 10, as -6 + 10 = 4 (b). As we both know, this is where it gets confusing, as the ones we are multiplying go like so:
..a.......c
+1.....-3
......X
+2.....+10
But, here is the key, do a diagram like this, illustrating which two you want to multiply. But, the ones that, when multiplied together, makes sense - are backwards. When you do this, draw lines from illustrating which ones make sense, like the above, with the 2 to the -3 and the 1 to the 10. Then, you do the opposite. You set it up so that it is 2 is with the 10 and 1 is with the -3, getting (2x +10)(x - 3)
Was that too complicated? I could illustrate the other way but its more complicated (but much less confusing - I never use the above method anymore, I use the complicated one) and I can't make vBulletin do square root signs. If this didn't help you, I'm sorry. Just keep in mind that a and c's two factors have to equal the middle number when multiplied together and added.
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