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to see if we have faith he knows but we have to prove it

There is no D on the diagram :/

Your question is worded really well .. I can see you have tried it properly for yourself.  I like that :)

7+- the square root of 9 divided by 2

I got it using the quadratic formula from x^2+4x-7=0

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-4 \pm \sqrt{4^2-4*1*-7} \over 2*1}\\ x = {-4 \pm \sqrt{16--28} \over 2}\\ x = {-4 \pm \sqrt{16+28} \over 2}\\ x = {-4 \pm \sqrt{44} \over 2}\\ x = {-4 \pm \sqrt{4*11} \over 2}\\ x = {-4 \pm2 \sqrt{11} \over 2}\\ x = {2(-2 \pm \sqrt{11} )\over 2}\\ x=-2 \pm \sqrt{11}$

So if you did not mistype your original equation then you did something wrong with the quadratic formula.

Now I will look at you ACTUAL question ... which is simpler :)

7+- the square root of 9 divided by 2

$x = {7 \pm \sqrt{9} \over 2}\\ x = {7 \pm 3 \over 2}\\ x = {7 + 3 \over 2} \qquad or \qquad x = {7 - 3 \over 2}\\ x = {10 \over 2} \qquad \;\; or \qquad x = {4 \over 2}\\ x = 5 \qquad \quad or\quad \quad x = 2\\$

This would have been the solution to      $x^2-7x+10=0$

64%=64/100=16/25

x=60/13

it's because i can't get it while i'm at school so i'm going to get it until i get home

*hands out muffins*

yeah i've been trying to get it but i have just not had enough time but i finally do

omg you r to sweet

i dont see what you mean

MUAHAHHAHAHHSAHHHAH

(Its a joke ... not to be taken seriously ...................................................................................................................................................................................................................................GEEZ

ACTUALLY depending on your religion some people think you live MANY times, a prime example being buddhism where they belive in reincarnation

As follows

$\lim_{x\rightarrow 3}\frac{1/x-1/3}{x-3}\rightarrow \lim_{x\rightarrow 3}\frac{3-x}{3x(x-3)} \rightarrow \lim_{x\rightarrow 3}\frac{-1}{3x} \rightarrow \frac{-1}{9}$