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1. Whats the best way to find the value of n by factorising?
3n^2 - 2n - 7400 = 0
2. (Original post by Nkhan)
Whats the best way to find the value of n by factorising?
3n^2 - 2n - 7400 = 0
That's one way, or the formula.
3. (Original post by Nkhan)
Whats the best way to find the value of n by factorising?
3n^2 - 2n - 7400 = 0
Note that when you plug n=50 into the equation, you find that the LHS=RHS=0. This means that (n-50) is a factor.

So you have:

By comparing both sides, you can see that:

and

Therefore B=148 so:

If you can't spot that factorisation immediately, then complete the square (or if you have calculator on hand, the quadratic formula).
4. multiply -7400 by 3 to get -22200
then find two numbers which have a difference of -2 and multiple to get -22200
so 148 and -150 :
3n^2 - 150n + 148n - 7400

look at the first part separately to the second part
3n^2 - 150n
take out the common factor of 3n
3n(n-50)

and do the same on the second part:
148(n-50) ---> both brackets show be equal

then you end up with 3n(n-50) + 148(n-50)
so (3n+148)(n-50)
this makes n= 50 and -49.333

hpe this helps!
5. multiply -7400 by 3 to get -22200
then find two numbers which have a difference of -2 and multiple to get -22200
so 148 and -150 :
3n^2 - 150n + 148n - 7400

look at the first part separately to the second part
3n^2 - 150n
take out the common factor of 3n
3n(n-50)

and do the same on the second part:
148(n-50) ---> both brackets show be equal

then you end up with 3n(n-50) + 148(n-50)
so (3n+148)(n-50)
this makes n= 50 and -49.333

hpe this helps!
6. Method for almost any quadratic equation is to find two numbers which...
- When multiplied together give the coefficient of the squared number multiplied by the added/subtracted number...
- And when added give the coefficient of the number on its own (technically to the power of one)

So as $Jellybeans$ said above you need to find two numbers which multiply to give -22200 and add to give 2. As the multiplied number is a negative you will know that one of these two numbers will be negative (confirmed by the very low -2n in contrast).
So with a bit of trial and error (i usually test a few numbers on some rough paper) you'll find the two numbers you need are 148 and -150 (multiply to give -22200 and add to give 2) and then you can follow through the answer above to get the final solution.

Alternatively if this method does not work (i.e. it is difficult to see an obvious solution) you can use the quadratic formula...

x = -b (+ or -) [squ root(b^2 - 4ac)]/[2a] <---- don't know how to do the symbols but if you type in quadratic formula into google it will come up for you.

Then you assign your a, b and c vals..
So...
a= 3
b = -2
c= -7400

and then put these all into the equation and work out for x (Don't forget to find the TWO roots (+ or -))
This method may take a while so to test there are solutions at all sub your a,b,c vals into the (-b^2 - 4ac). If the outcome is negative, you know there are no solutions.
7. (Original post by Farhan.Hanif93)
Note that when you plug n=50 into the equation, you find that the LHS=RHS=0. This means that (n-50) is a factor.
What would prompt you to check n=50?
8. (Original post by ghostwalker)
What would prompt you to check n=50?
Well it was a guess. I knew that if I'm looking for a positive root then 3 times the square of that root must be larger than 7400. From there, n=50 was an obvious first choice. Also following that, the dead giveaway was that 50 is a factor of 7400.
9. (Original post by ghostwalker)
What would prompt you to check n=50?
I guess it's just intuitive, try educated guesses
10. (Original post by ghostwalker)
What would prompt you to check n=50?
I'd say the -2 in the middle is roughly zero so one of our numbers is roughly 3 times the other. Since their product is 7400 on of them is roughly so I'd say one of them is about 50.

This works quite well in these cases.

Edit. LaTeX typo.
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