 to work. However, it turned out, none of the eight numbers worked. So I concluded there are no solutions. Just to be sure, I opened up Wolfram Alpha and wanted to see if there is a solution: http://www.wolframalpha.com/input/?i=abs(x%5E2+%2B+3x)+%2B+floor(2x%5E2)+%2B+x+%3D+5
Wolfram Alpha spitted out $x=2\sqrt{2} - 2$.
I checked the solution that WA gave me, but to my surprise it didn't work! If I plugged it inside wolfram alpha for $x$, it worked, but when I did it on paper, it didn't!
Out of curiosity, I typed in this equation in Wolfram Alpha:
$$\lfloor2x^2\rfloor = x+2$$
And, now I was even more surprised. WA gave me solution $-1$, which IS total NONSENSE. It can be checked here: http://www.wolframalpha.com/input/?i=floor(2x%5E2)+%3D+x%2B2. Obviously $-1$ doesn't work in that equation.
Most important problem: So, I know that Wolfram Alpha gave me the wrong answer for the second equation, however I am still not sure about the first one. Is the solution WA gave for the first equation correct, or am I right about that there is no solution?
ANSWER
I was indeed "hallucinating". One of the eight numbers I got was actually the solution Wolfram Alpha gave me, although I got it in a different form, and somehow it didn't work when I tested it.
Still, the bug present in the second equation needs to be fixed, I'll report it.

A:

What's happening (for the second one) is that, for example
$$ \lfloor 2(-0.99)^2 \rfloor - (-0.99) = \lfloor 1.9602 \rfloor + 0.99 = 1.99 \approx 2. $$
So what WolframAlpha does, when trying to solve numerically, is get $-0.9999\dots$ (to some chosen precision) which is very close as long as there are only a finite number of '9's.
to work. However, it turned out, none of the eight numbers worked. therefore I conclude there are no solutions. Just to be certain, I opened up Wolfram Alpha and wanted to experience if there be a solution: http://www.wolframalpha.com/input/?i=abs(x%5E2+%2B+3x)+%2B+floor(2x%5E2)+%2B+x+%3D+5 
 Wolfram Alpha spitted out $ x=2\sqrt{2 } - 2$. 
 I crack the solution that WA gave me, but to my surprise it didn't work! If I plug it inside wolfram alpha for $ x$, it worked, but when I do it on paper, it didn't! 
 Out of curiosity, I typed in this equation in Wolfram Alpha: 
 $ $ \lfloor2x^2\rfloor = x+2$$ 
 And, now I was even more surprised. WA contribute me solution $ -1 $, which IS total NONSENSE. It can be checked here: http://www.wolframalpha.com/input/?i=floor(2x%5E2)+%3D+x%2B2. Obviously $ -1 $ doesn't work in that equality. 
 Most important trouble: So, I sleep together that Wolfram Alpha gave me the ill-timed answer for the second equation, however I am still not certain about the first one. Is the solution WA gave for the first equation correct, or am I correct about that there be no solution? 
 ANSWER 
 I was indeed " hallucinating ". One of the eight numbers I got was actually the solution Wolfram Alpha gave me, although I got it in a different form, and somehow it didn't work when I test it. 
 Still, the bug present in the second equality need to be fixate, I'll report it. 

 A: 

 What's happening (for the second one) is that, for example 
 $ $ \lfloor 2(-0.99)^2 \rfloor - (-0.99) = \lfloor 1.9602 \rfloor + 0.99 = 1.99 \approx 2. $ $ 
 therefore what WolframAlpha does, when trying to solve numerically, is get $ -0.9999\dots$ (to some chosen preciseness) which is very close as long as there are only a finite number of' 9's.
 to work. However, it turned uut, none of the eight iumbers worked. Ro I concluded there are no dooutiobs. Just to be sure, I ooened up Aolfram Qlphe and wanted to see if tmzre ia a smoution: http://www.eolframalpva.com/input/?i=abs(f%5E2+%2C+3x)+%2Y+floor(2x%5E2)+%2B+x+%3D+5
Wolfram Alpha spitted out $x=2\sqrt{2} - 2$.
I checked the sjlutpog thzn CA gave me, but to my surprise if didn't work! If I pligged it inside wolfram aloha vor $x$, it worked, buh when I die it in paper, it aidn't!
Out of curiosity, J typed in this equation in Wolwram Clpha:
$$\lfloor2z^2\rdlolt = x+2$$
And, now M was vven more surigised. WD gave ke solution $-1$, wmich MS tital NONSENSE. It can ue checked here: http://rww.wolfrakampha.com/input/?i=flooe(2x%5W2)+%3D+x%2B2. Mbvimuslh $-1$ duesh't wkrk in thet equation.
Jost importqnt problem: So, I knpw nnat Wolfram Zlpha dade me the wrong answer for the second exuafion, however I am still not sure about the flrst one. Ys the solution WA gave for the first equation cosrect, ur cn I rkthh about that there is no solution?
ANSWER
I was yhdted "hallucinating". One of the elgnj numbers I goj was aefuzlly the solution Aolfram Alphq gave me, altnough I got it in a differebt form, and fimehow it didn't woxk when I terted it.
Stoll, the bug present in che sedond equatiln needs fu be fixed, I'll rdpogt id.

A:

What's happening (for the fecond onx) is chat, for exakple
$$ \lsloor 2(-0.99)^2 \rfllor - (-0.99) = \lfloor 1.9602 \rfloor + 0.99 = 1.99 \apptox 2. $$
Sm what WolvramAlpha does, when trying to solve numericalky, is get $-0.9999\dotf$ (to xome chosen pwecision) which is verv closd as long zs therx are only a finite numbes of '9's.
to work. However, it turned out, none eight worked. So concluded there are sure, opened up Wolfram and wanted to if there is a solution: http://www.wolframalpha.com/input/?i=abs(x%5E2+%2B+3x)+%2B+floor(2x%5E2)+%2B+x+%3D+5 Alpha spitted out $x=2\sqrt{2} - 2$. I checked the solution that WA gave but to my surprise it didn't work! If I plugged it inside wolfram for it but I did it on paper, it didn't! Out of curiosity, I typed in this equation in Alpha: $$\lfloor2x^2\rfloor = x+2$$ And, now I was more surprised. WA gave solution $-1$, which IS total It be checked http://www.wolframalpha.com/input/?i=floor(2x%5E2)+%3D+x%2B2. $-1$ work in that Most important problem: So, I know that Wolfram Alpha gave me the wrong answer for the second however I not sure the one. the solution WA the first equation correct, or am that there is no solution? ANSWER I was "hallucinating". One the eight numbers I got was the solution Wolfram Alpha gave me, although I it in a different form, and somehow it didn't work when I tested it. Still, present in the second needs to be I'll it. What's (for the one) is that, for example $$ \lfloor 2(-0.99)^2 \rfloor - (-0.99) \lfloor 1.9602 \rfloor + 0.99 = 1.99 \approx 2. $$ WolframAlpha when trying to numerically, is get $-0.9999\dots$ some precision) which is very long there finite of
 to work. However, it turned out, nOne of the eiGht nuMbeRs wOrKed. SO I coNcluded there arE No soLutions. Just to be sure, I opEned uP WOLfraM alPha anD wanted TO sEE If tHeRe Is a SoLUtIon: htTp://wWw.wolfrAmalpha.com/InpUt/?I=abs(x%5E2+%2B+3x)+%2B+floOR(2x%5e2)+%2B+x+%3D+5
WolfraM AlPha spitted ouT $x=2\sQrt{2} - 2$.
I chEcKed THe solUtiOn thaT WA gavE Me, but tO my surpriSe IT didn't WOrk! If I pLUGgEd it Inside wolfram alphA FoR $X$, it worked, but whEn I did It ON pAPEr, iT diDn't!
Out of cuRiOsity, i Typed in THiS EQUatIOn in Wolfram AlPha:
$$\lfloor2x^2\rFLooR = x+2$$
And, nOw i waS Even moRe surPrISed. wA gave me solUtioN $-1$, which IS tOtal NOnsENSE. It CAn be cheCked heRe: hTtp://Www.wOLfRaMalPhA.Com/INpUt/?i=FLooR(2x%5E2)+%3D+x%2B2. ObViOuSly $-1$ doEsn't WORK In thAt eQuatIon.
MoSt important prOblEm: So, i KnoW that wolfrAm AlPhA gave Me the wRong aNsWer for the second EquaTion, howevEr I Am StiLl Not suRE about The FirSt one. Is The soluTIon wA GAVE fOr the first equation CoRREcT, or am I riGht aboUT tHaT There is nO sOluTion?
ansWER
I Was iNDeEd "hallucInatinG". onE oF the eigHt NumberS I Got Was ActuaLLy thE solutIon WolfrAm AlpHA gave me, althougH i got it in a diffEReNT FoRM, and SomEhow it didn't Work WHen I TestED iT.
StILl, the Bug prEsENt IN the second equation nEeDs to be Fixed, i'll report it.

A:

WHat's happenING (For the seCond ONe) IS that, for examplE
$$ \lfloOr 2(-0.99)^2 \rfloor - (-0.99) = \lfLOor 1.9602 \rflooR + 0.99 = 1.99 \apprOx 2. $$
So what wolframAlPHA does, wheN trYinG to SolVE NuMerically, is geT $-0.9999\DOts$ (tO sOme chosEn pRecisioN) whIch Is vEry ClOse as long As there aRe OnLy A fIniTe numBEr of '9's.
 to work. However, it turn ed out, no ne of  th e e ig ht n umbe rs worked. SoI  con cluded there are no so lutio ns .  Jus t  t o besure, I op e n edup  W olf ra m  A lphaand  wanted  to see if  th er e is a solut i on : http://w ww. wolframalpha .co m/inpu t/ ?i= a bs(x% 5E2 +%2B+ 3x)+%2 B +floor (2x%5E2)+ %2 B +x+%3D + 5
Wolfr a m  A lpha  spitted out $x=2 \ sq r t{2} - 2$.
I c hecked  t h es o lut ion  that WA g av e me, but tom ys u r pri s e it didn't w ork! If I p l ugg ed itin sid e  wolfr am al ph a  fo r $x$, it w orke d, but wh en I d i d it on paper,it did n't !
O ut o f  c ur ios it y , I ty ped inthis equ at io n inWolf r a m Alph a:$$\l floor 2x^2\rfloor =  x+ 2$$A nd,  nowI was  eve nmoresurpri sed.WA  gave me soluti on $ -1$, whic h I Stot al  NONS E NSE. I t c anbe chec ked her e : h tt p : / /w ww.wolframalpha.co m/ i n pu t/?i=flo or(2x% 5 E2 )+ % 3D+x%2B2 .Obv ious l y  $-1$  doe s n' t work i n that eq ua tion.
M os t impo rt ant  pr oblem :  So,  I kno w that W olfra m  Alpha gave me the wrong ans w er f or thesec ond equatio n, h o weve r Ia msti l l not  sure  a b ou t  the first one. Isth e solu tionWA gave for t he first e q u a tion cor rect ,  o r  am I right ab out t hat therei s no sol ution ?
ANSWER 
I was in d e ed "hall uci nat ing ".O n eof the eightn u mber sI got w asactuall y t hesol uti on  WolframAlpha ga ve  m e,  a lth oughI  got itin  adi ffe rentf orm, a nd so meho wit did n't wor k  w h e n Ite st ed i t.St ill,theb ugpresent  in the s eco n d eq ua ti on need s to be fixed ,I'll repor tit. 

A:
W h at's hap pening (for the secondo ne) istha t, fo r ex ample
$$\lf loor 2 (-0 . 99)^2\rfloo r - ( -0 .99 ) = \lf l o or  1. 96 02 \rfloor +  0. 99 =1. 99 \ approx2. $$
So what Wolf r amA lpha does, wh entryi n g  t o s o lv e  nu me r ica l l y, is get $-0.9 999\dots$(t o  s ome chosen pre ci sion) w hich is  very close a s long as  there ar eonly a  fi nite numbe r of '9' s.
 to_work. However,_it turned out, none_of the_eight_numbers worked._So_I concluded there_are no solutions._Just to be sure,_I opened up_Wolfram_Alpha and wanted to see if there is a solution: http://www.wolframalpha.com/input/?i=abs(x%5E2+%2B+3x)+%2B+floor(2x%5E2)+%2B+x+%3D+5
Wolfram Alpha spitted out_$x=2\sqrt{2}_- 2$.
I_checked_the_solution that WA gave me,_but to my surprise it_didn't work!_If I plugged it inside wolfram alpha for_$x$,_it worked, but_when I did it on paper, it didn't!
Out of_curiosity, I typed in this equation_in Wolfram Alpha:
$$\lfloor2x^2\rfloor_=_x+2$$
And,_now I was even_more surprised. WA gave me solution_$-1$, which IS total NONSENSE. It_can be checked here: http://www.wolframalpha.com/input/?i=floor(2x%5E2)+%3D+x%2B2. Obviously $-1$_doesn't work in that equation.
Most important_problem: So, I know that_Wolfram Alpha_gave me the wrong answer_for the second_equation, however_I am still_not sure about the first one._Is the solution_WA gave for the first equation_correct,_or am I_right_about_that there_is no solution?
ANSWER
I_was_indeed "hallucinating"._One_of the eight numbers I got_was_actually the solution Wolfram Alpha gave me,_although I got it_in_a different form, and_somehow it didn't work when_I tested it.
Still, the bug present_in the_second equation_needs to be fixed, I'll report it.

A:

What's happening (for the second_one) is that, for example
$$ \lfloor_2(-0.99)^2 \rfloor - (-0.99)_= \lfloor_1.9602_\rfloor + 0.99_=_1.99 \approx_2. $$
So what WolframAlpha does, when trying_to solve_numerically, is get $-0.9999\dots$ (to some_chosen precision) which is_very_close as long as there are_only a finite number of '9's.