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## 2n^2+4n-16=0

Hi Raven,

2n^2 + 4n -16 = 0

This is standard form aN^2 + bN + c    (a=2,b=4,c= -16)

1)examine the possible factors of the a & c constants [ c=(-2)(8) or (2)(-8) or (4)(-4) or (1)(-16) or (-1)(16) ]
[a = (1)(2) ]
2)look at the possible combinations and try to construct the 'b' constant from the candidate factors.  For example consider the form

(2n +  Q)(n + R)

QR = c = -16
(Q + 2R) = b = 4

After a little study, you'll see that (Q=8 , R=-2) yields the correct 'b' So we FOUND the solution just by examination.  This doesn't work all the time, but is faster if the roots are simple

(2n + 8)(n - 2)-----> n=-4, n=2   are the solutions

2)If you cannot find simple factors, you are left with the quadratic formula which always provides the correct answer:  n = [ -b ± sqrt(b^2 - 4ac) ]/2a

n= [ -4 ± sqrt(16 - 4(2)(-16) ]/2(2)
n= -1 ± sqrt(16 + 128)/4
n= -1 ± sqrt(144)/4   (sqrt(144)=12)
n= -1 ± 3
n= -4, 2     (Same as we got above)
Hi Raven;
2n2+4n-16=0
For the FOIL...
FIRST must be (2x)(n)=2n2.
OUTER and INNER must add-up to 4n.
LAST must be (16)(1) or (1)(16) or (8)(2) or (2)(8) or (4)(4) and one number must be negative to produce -16.
(2n+8)(n-2)=0
Let's FOIL...
FIRST...(2n)(n)=2n2
OUTER...(2n)(-2)=-4n
INNER...(8)(n)=8n
LAST...(8)(-2)=-16
2n2-4n+8n-16=0
2n2+4n-16=0
(2n+8)(n-2)=0
Either or both parenthetical equations must equal zero.
2n+8=0
2n=-8
n=-4

n-2=0
n=2

First, divide both sides by 2. You will get:

n2+2n-8=0

Now either factor out or use quadratic formula.

1) Factoring.
Product of roots shall be equal to -8. Sum of them equals to -2. Obvious choice is -4 and 2. (-4)*2=-8, -4+2=-2.

So we have the following factoring: (n+4)(n-2)=0. This means that either n=2 or n=-4.

n1,2={-2±√[(-2)2-4*1*(-8)]}/2
n1=[-2+√36]/2=(-2+6)/2=2
n2=[-2-√36]/2=(-2-6)/2=-4

the formula should be ax^2+bx++c=0
Hi, Raven:

What you wrote is called quadratic equation, not quadratic formula. The second thing, you shall think generally, not in specifics. If I write yx2+zx+s=0, this is still quadratic equation. And m*sin2(x)+h*sin(x)+w=0 is quadratic equation, with respect to sin(x), of course.
Look at what Kirill did:divided both sides of the equation by 2 to get n^2+2n-8=0.
Another way to look at the problem now is that we want two numbers whose product is 8(disregarding signs for the moment). Because 2 is even the two factors for 8 must both be even which eliminates 1 and 8 and leaves us with only 2 and 4.
4*(-2)=-8 and 4+(-2)=2
Always factor out any numerical common factors and make sure it's the GCF. This will make your work simpler. Always put your polynomial (equation) in the form ax^2+bx+c(=0).
Look at a,b, and c. Each one is either even(E) or odd(O). There are 8 possibilities for a,b, and c:
OOO
EEE
EEO
OEE
EOE
OOE
EOO
OEO
There are rules for 7 of the 8 possibilities.