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November 28, 2014

November 28, 2014

Posted by **John** on Thursday, May 15, 2008 at 6:13pm.

-3x to the second+18x+12

-5x to the second-13x=6

-5x to the second-6x=1

-and xto the second + 6x=10

I can't use the quadratic equation. I have to use the a times c method.

- RE: Math -
**Reiny**, Thursday, May 15, 2008 at 7:30pmTwo things:

1. I think your first equation has a typo, since it is not an equation

2. I think you are using the - in front of each equation not as a negative sign, but rather as a dash. Why????

So I will assume that your second equation is

5x^2 - 13x = 6 (the ^ means 'to the exponent')

5x^2 - 13x - 6 = 0

(a)(c) = 5(-6) = -30

now look for factors of -30 which have a sum of -13

that would be -15 and 2

check :(-15)(2) = -30, -15 + 2 = -13

now replace the middle term of -13x with -15x+2x

5x^2 - 15x + 2x - 6 = 0

take out a common factor from the first two terms, and from the last two terms

5x(x-3) + 2(x-3) = 0

now x-3 is a common factor, so ...

(x-3)(5x+2) = 0

x-3 = 0 or 5x+2 = 0

x = 3 or x = -2/5

- RE: Math -
**Reiny**, Thursday, May 15, 2008 at 7:31pmDo the others the same way

- RE: Math -
- RE: Math -
**John**, Thursday, May 15, 2008 at 9:13pmSo You using the quadratic or a time c method will give me the same answer?

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