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Thanks 

who are you? Are you the guest who answers every question right? How are you so good? Maybe you should make an account 

It's okay ;)

Let Aiden have \(x\) games, and so then Will has \(\frac{x}{8}\). We then have the equation

\(x=\frac{x}{8}+168\)

Moving the \(\frac{x}{8}\) to the left, the equation turns into

\(\frac{7x}{8}=168\)

Therefore, Aiden has \(x=192\) games.

This answer is not helpful

sorry

Nice one Asinus!!!!

please stop. It's very annoying to just post about random things. Just look at the sticky topics.

can you remove the $ sign?

$the \ answer \ is \ incorrect, \ face \ it.$

Hello Guest!

I'm sorry, but there's nothing that can be done.

  !

Using the quadratic formula gives:

\(_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:7\left(-5\right)}}{2\cdot \:7}\)

\(x_1=\frac{-1+\sqrt{141}}{2\cdot \:7},\:x_2=\frac{-1-\sqrt{141}}{2\cdot \:7}\)

\(x=\frac{-1+\sqrt{141}}{14},\:x=\frac{-1-\sqrt{141}}{14}\)

So if a and b are roots:

\((\frac{-1+\sqrt{141}}{14}-4)(\frac{-1-\sqrt{141}}{14}-4)\)

(EDIT:  Just you are calc if you don't know how to solve!!!)

Just solve that and you got the answer 

-Vinculum

He didn't capitalise the letter p in this name lol

I'm pretty sure CPhill would say "CPhill" instead of "Cphill"

Also, CPhill never puts his name at the end of his posts just the  

Edit: I double checked in Wolfram|Alpha.

It should be right.

https://www.wolframalpha.com/input?i2d=true&i=Power%5B%5Cleft%5C%2840%292y-7%5Cright%5C%2841%29%2C5%5D

Sorry Bro lemme check again

You mean 3+7+14+50? not 3+1+14+50...

-Vinculum

Edit : 3+7+14+50

its 74

Hello, 

\(x^2-y^2=51+xy\)

\(x^2-y^2-xy=51\)

\(x^2-yx-y^2-51=0\)

Solve with the quadratic formula:

\(x_{1,\:2}=\frac{-\left(-y\right)\pm \sqrt{\left(-y\right)^2-4\cdot \:1\cdot \left(-y^2-51\right)}}{2\cdot \:1}\)

\(x_{1,\:2}=\frac{-\left(-y\right)\pm \sqrt{5y^2+204}}{2\cdot \:1}\).

I think you got \(y\)...

Let me know if you need help becoz this might get very messy.Good luck

True I guess

I think there was something wrong with the binomial expansion

No I was the asker and your answer was incorrect, chill.

I mean it could be him just not signed in :shrug:

Think again, Vinculum