1. ## factoring

Does this solve to a real number?

10x^2+9x-9=0

2. Why factor? The quadratic formula will answer that for you. You don't even need to use the entire formula!

3. Yeah, the quadratic formula will work and is simpler to explain. As far as factoring goes, here is what to do.

10x^2+9x-9=0

First you want to find what two numbers who's sum is 9 and when multiplied equal -90. Those two numbers are -6 and 15.

Then the problem works out as such...

(10x^2-6x) + (15x-9)

2x(5x-3) + 3(5x-3)

Therefore, it should factor out to (2x+3)(5x-3)=0. Your answers should be x = -(3/2) or (3/5).

4. Originally Posted by galanm
Does this solve to a real number?

10x^2+9x-9=0
The only restriction to be a single or a pair of real numbers; if you write a general equation $a\cdot{x^2}+b\cdot{x}+c=0$ and $\sqrt{b^2-4\cdot{ac}}\geq{0}$ then your solution is a pair of different real numbers if is strictly large than zero or a single real number if it's equal to zero

5. Originally Posted by some_nerdy_guy
Yeah, the quadratic formula will work and is simpler to explain. As far as factoring goes, here is what to do.

10x^2+9x-9=0

First you want to find what two numbers who's sum is 9 and when multiplied equal -90.
No, they must multiply to -9, not -90.

Those two numbers are -6 and 15.

Then the problem works out as such...

(10x^2-6x) + (15x-9)

2x(5x-3) + 3(5x-3)

Therefore, it should factor out to (2x+3)(5x-3)=0. Your answers should be x = -(3/2) or (3/5).
You've solved the wrong problem!

6. Huh? It's not wrong, I plugged in the answers and they checked out.

10x^2+9x-9=0

ax^2+b+c=0

You multiply 'a' and 'c' which would be 10 and -9 and will equal -90. Then you see what two numbers when added together will have a sum of 'b' and when multiplied together will equate 'ac' (in this case, -90). How are you concluding that they should equal -9?