Does this solve to a real number?
10x^2+9x-9=0
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).
The only restriction to be a single or a pair of real numbers; if you write a general equation $\displaystyle a\cdot{x^2}+b\cdot{x}+c=0$ and $\displaystyle \sqrt{b^2-4\cdot{ac}}\geq{0}$ 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
No, they must multiply to -9, not -90.
You've solved the wrong problem!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).
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?