how to tranform into quadratic form and solve for the unknown?

>>351185
>i'm just desperate to pass
Should've studied desperately then. This is literally the baby's first math, if you can't solve this you're better off failing and taking the class again.
http://www.virtualnerd.com/algebra-2/polynomials/equations/solve-equations/quadratic-form-definition
https://www.cliffsnotes.com/study-guides/algebra/algebra-ii/quadratics-in-one-variable/solving-equations-in-quadratic-form

>>351202
you might want to actually look at the problems kid

>>351185
For the first 2, you need to make n = m^2, apply the quadratic formula and then take the square root for each answer. For the last one you can't do this and there are 4 answers. To solve this you need to convert it to a form where

(am^2 + bm +c)(dm^2 + em + f) = 0

Your coefficients in decreasing order are
[16,8,-27,-7,6]

a*d = 16
a*e + b*d = 8
a*f + b*e + c*d = -27
b*f + c*e = -7
c*f = 6

Wow, that's a lot of crap to do. The good news is that you can make something up for a and d, do the same for c and f, and then use them to determine b and e. Plug everything in at the end to make sure it checks out - if not, you'll have to re-do one of your initial guesses. After which, you can take either parenthetical term I described above, and do the quadratic equation to get 2 answers for each. Because for the whole thing to be 0, only one of those terms has to be 0. In total, you should get 4 answers with no calculator required.

>>351240
I think there's a mathematical way to figure out a-f, but it's very complicated and probably out of the class's scope. I found that making the easiest initial guess led to

a = d = 4 (my guess)
b = e = 1
4f + 1 + 4c = -27 ---> 4(f + c) = -28
f + (6/f) = -7 since c = 6/f. At this point you can guess, but just to be analytical:
f(f + (6/f) + 7) = 0
f^2 + 7f + 6 = 0
Quadratic formula on this says that
f = (-7 plus or minus sqrt(25))/2 = -6 or -1
If we make f = -6, then c = -1 and vice-versa.

I made that post, then felt bad because I didn't try solving the equation myself to verify that it's doable analytically. I was still wrong by telling you to guess for 4 different variables, when only 2 is necessary.

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>>351240
>>351246
Wow that's awesome, gj dude.

>>351281
No prob

>>351185
you goobers overcomplicated (3) severely
set (4m^2+m) = x
(3) becomes
x^2 - 7x + 6 = 0
(x - 6)(x - 1) = 0
then put it back
[(4m^2 + m) - 6][(4m^2 + m) - 1] = 0
(4m^2 + m - 6)(4m^2 + m -1) = 0
boom, quadratics
no guesswork required

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>>351529
I figured there had to be an easier way...