fun ^ 10 x int ^ 40 = Ir2

How do I solve this equation?

\sum_{i=0}^nx_i

$\sum_{i=0}^nx_i$

$$\sum_{i=0}^nx_i$$

[\newcommand{\Bold}[1]{\mathbf{#1}}\left[\mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{1}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{1}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{3}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{1}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{\left(i + 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{3}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{7}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{2}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{9}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{i \, 2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{11}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{3}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{13}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{7}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{\left(i - 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{4}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{17}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{9}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{19}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{19}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{9}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{17}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{4}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{\left(i + 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{7}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{13}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{3}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{11}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{i \, 2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{9}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{2}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{7}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{3}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{\left(i - 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{1}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{3}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{1}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{1}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}\right]\]

\left[\mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{1}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{1}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{3}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{1}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{\left(i + 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{3}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{7}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{2}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{9}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{i \, 2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{11}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{3}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{13}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{7}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{\left(i - 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{4}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{17}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{9}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(\frac{19}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{19}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{9}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{17}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{4}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{\left(i + 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{7}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{13}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{3}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{11}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{i \, 2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{9}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{2}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{7}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{3}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = -\frac{\left(i - 1\right) \, 2^{\left(\frac{21}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{2 \, \mbox{fun}^{\left(\frac{1}{4}\right)}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{1}{5} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{3}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{1}{10} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)} e^{\left(-\frac{1}{20} i \, \pi\right)}}{\mbox{fun}^{\frac{1}{4}}}, \mbox{int} = \frac{2^{\left(\frac{1}{40}\right)} \mbox{Ir}^{\left(\frac{1}{40}\right)}}{\mbox{fun}^{\frac{1}{4}}}\right]

hey i dont think I in the Ir2 is integer rather it would be inverted reality. and the 2 would be the combining of the inverted reality and the true reality.

Obviously, Ir stands for Iridium. Which is element number 77.

その目誰の目？

http://www.thewebcomiclist.com/o/8538

$x_I$

[math]x_i

[math]x_i[math]

fgxd

Dear diary, Today OP was a fag.

Kill me

Kill me

while having fun with a heighten intelligence you may blur the realms of your imagination with reality provided if the collective conscious see what you see.

simple version: if a smart man play a prank and said that pineapples are apples everyone around see it as true if they are less intelligent.

>>20
Solved like a true physicist.

compounded of two bodies, viz. sol and Luna; is puffed
up, swells, putrefies, is raised up, and does increase by
the receiving from the animated nature and substance.
Our water, or vinegar as aforementioned, is the vinegar of
the mountains, i.e. of sol and Luna; and therefore it is
mixed with gold and silver, and sticks close to them
perpetually; and the body receives from this water a white
tincture, and shines with inestimable brightness. Who so
knows how to convert, or change the body into a medicinal
white gold, may easily, by the same white gold, change all
imperfect metals into the best or finest silver. And this
white gold is called b the philosophers "luna alba
philosophorum, argentum vivum, album fixum, aurum
alchymiae, fumus albus" and therefore without this our
antimonial vinegar the stone of the philosophers cannot be
made. The reason, is because in our vinegar there is a
double substance of argentum vivum, the one from
antimony, and the other from mercury sublimated, it does
give a double weight and substance of fixed argent vive,
and also augments therein the native colour, weight,
substance and tincture thereof.

compounded of two bodies, viz. sol and Luna; is puffed
up, swells, putrefies, is raised up, and does increase by
the receiving from the animated nature and substance.
Our water, or vinegar as aforementioned, is the vinegar of
the mountains, i.e. of sol and Luna; and therefore it is
mixed with gold and silver, and sticks close to them
perpetually; and the body receives from this water a white
tincture, and shines with inestimable brightness. Who so
knows how to convert, or change the body into a medicinal
white gold, may easily, by the same white gold, change all
imperfect metals into the best or finest silver. And this
white gold is called b the philosophers "luna alba
philosophorum, argentum vivum, album fixum, aurum
alchymiae, fumus albus" and therefore without this our
antimonial vinegar the stone of the philosophers cannot be
made. The reason, is because in our vinegar there is a
double substance of argentum vivum, the one from
antimony, and the other from mercury sublimated, it does
give a double weight and substance of fixed argent vive,
and also augments therein the native colour, weight,
substance and tincture thereof.

I.. think its wierd..

You have been into chaos head soo much watch new anime man!

what

i like pie

Thomas Covenant. The White Gold Wielder

その目誰の目

What the shit is this? Honestly fun ^ 10 x int ^ 40 = Ir2 means
fun ^ 10 =10fun
int. ^ 40 = your iq is 40 ( your retarted)
ir2 is just the interior raito of your brain
so 10 things are fun to you because your brain is small and can't handle more. Duh

guise, i think Ir2 could mean lierally:
I are two (years old)

This is the incantaion that kind Gladial will use to plunge this world into Chaos. Quit wasting time, Takumi, find your Di-sword. If you don't the Black Knights will not be able to stop what has been set into motion.

その目だれの目
how it's written in the game. i don't see 誰 anywhere

Pie suki da ta.

It could only be solved if you had an IBN 5100. Go to akihabara and find one, I hear this dude who looks like a chick has one. Ruki or something.

Hello

Assume a Ding-a-Ling, if you have it not.

その目だれの目？

>>72
Hey, you, The Neidthart! I know it's you, you can not trick me! Dare you not to touch my IBN5100! Wait, are you cooperating with "Organization"? Damn, I should have guessed... But you can not scare me with such a bullshit! Who do you think I am? I'm the insane mad scientist Hououin Kyouma!
Go find your own computer, or I will release my right hand... I'm sure you don't want that to happen!

fun might be function the ^ stands fore multiplucation the x is a verible int is internet the ir is a persons veuw on realety the 2 is dubling it so all you have to do is find x and you would hav theretical mind altration using some sortive of brodcasting sistum lick cp or radio p.s whare did you come up with this
whose eye are those are?

sono me dare no me?

This equation is not one that will be solved for at least the next 50 years, don't worry about it until then.

あぁ、駄目だ駄目だ,全然駄目だ!