How to write Quadratic equation with negative coefficient in fp

For a=1, b=-5, c= 6, $ca x^2+cb x + cc=0$ gives 1x^2 + -5x + 6

But i want to have x^2 -5x + 6

documentclassbeamer
usepackagefp
begindocument
beginframeQuadratic equation
FPsetca1
FPsetcb-5
FPsetcc6
FPqsolvexonextwocacbcc
FPevalxoneclip(round(xone:4))
FPevalxtwoclip(round(xtwo:4))
Quadratic equation : $ca x^2+cb x + cc=0$\[1cm]
Result: $x = xone quad textand quad x = xtwo$
endframe
enddocument

Some comparison are necessary. This assumes the coefficients are integers.

documentclassbeamer
usepackagefp

newcommandquadratic[4][x]%
 FPsetca#2%
 FPsetcb#3%
 FPsetcc#4%
 FPqsolvexonextwocacbcc
 FPevalxoneclip(round(xone:4))%
 FPevalxtwoclip(round(xtwo:4))%
 Quadratic equation: $
 ifnumca=1
 else
 ifnumca=-1
 -%
 else
 ca
 fi
 fi
 #1^2%
 ifnumcb=0
 else
 ifnumcb>0
 +%
 ifnumcb=1
 else
 cb
 fi
 else
 ifnumcb=-1
 -%
 else
 cb
 fi
 fi
 #1%
 fi
 ifnumcc=0
 else
 ifnumcc>0
 +
 fi
 cc
 fi
 $\[bigskipamount]
 Result: $#1=xone$ and $#1=xtwo$%


begindocument
beginframeQuadratic equation

quadratic1-56

bigskip

quadratic[t]231

bigskip

quadratic20-8

endframe
enddocument

With expl3:

documentclassbeamer
usepackagexparse

ExplSyntaxOn

NewDocumentCommandquadraticOxmmm
 
 Quadratic~equation:~$
 str_case:nnF #2 
 
 1
 -1-
 
 #2
 #1^2
 str_case:nnF #3 
 
 0
 1+#1
 -1-#1
 
 fp_compare:nT #3>0 + #3#1 
 fp_compare:nF #4 = 0 
 
 fp_compare:nT #4 > 0 + 
 
 #4
 $\[bigskipamount]
 Result:~$#1=sandu_solve:nnnn+#2#3#4$~and~
 $#1=sandu_solve:nnnn-#2#3#4$

cs_new:Nn sandu_solve:nnnn
 
 fp_eval:n round( ( -(#3) #1 sqrt((#3)^2-4*(#2)*(#4)) )/(2*(#2)), 4) 
 
ExplSyntaxOff

begindocument
beginframeQuadratic equation

quadratic1-56

bigskip

quadratic[t]231

bigskip

quadratic20-8

endframe
enddocument

Will also work with addterm -5x in addition to the intended addtermcb x.

The addterm macro takes a single argument, expands it once, and passes it to addtermaux. The addtermaux definition will grab the first token of the argument and examine to see if it is a minus - character. If so, it typesets a - and the rest of the argument. If not, it sees whether the first token was a + character. If so, it typesets a + and the rest of the argument. If neither of the above cases apply, it typesets a + and the complete argument.

In this way, the right output is provided whether cc is set to 6 or set to +6.

documentclassbeamer
usepackagefp
newcommandaddterm[1]expandafteraddtermaux#1relax
defaddtermaux#1#2relaxifx-#1-#2elseifx+#1+#2else+#1#2fifi
begindocument
beginframeQuadratic equation
FPsetca1
FPsetcb-5
FPsetcc6
FPqsolvexonextwocacbcc
FPevalxoneclip(round(xone:4))
FPevalxtwoclip(round(xtwo:4))
Quadratic equation : $ca x^2 addtermcb x addtermcc=0$\[1cm]
Result: $x = xone quad textand quad x = xtwo$
endframe
enddocument

Edit: See below an improved version.

Note the [fragile] in beginframe. Necessary with FPifpos.

documentclassbeamer
usepackagefp
begindocument
beginframe[fragile]Quadratic equation
FPsetca1
FPsetcb-5
FPsetcc6
FPqsolvexonextwocacbcc
FPevalxoneclip(round(xone:4))
FPevalxtwoclip(round(xtwo:4))
FPevalbabsclip(round(abs(cb):4))
FPevalcabsclip(round(abs(cc):4))

Quadratic equation : $ca x^2$ FPifposcb $+$ else $-$ fi $babs x$ FPifposcc $+$ else $-$ fi $cabs=0$ %\[1cm]

Result: $x = xone quad textand quad x = xtwo$


endframe
enddocument

Improved version

This version handle better special situations (when some coefficients of the equation became -1, 1 or 0).

As fp's FPqsolve doesn't handle equations without solutions (it emit an error), my code don't display correctly equations where the "x^2" AND the "x" term are null (it display, when the compiler don't stop at errors, something like: Quadratic equation: +6 = 0). This code is intended to be used only when the equation has real(s) solution(s).

documentclassbeamer
usepackagefp
begindocument
beginframe[fragile]Quadratic equation
FPsetca1
FPsetcb-5
FPsetcc6
FPqsolvexonextwocacbcc
FPevalxoneclip(round(xone:4))
FPevalxtwoclip(round(xtwo:4))
FPevalaabsclip(round(abs(ca):4))
FPevalbabsclip(round(abs(cb):4))
FPevalcabsclip(round(abs(cc):4))
newcommandsignaFPifnegca -elsefi
newcommandpositiveSignBWithAFPifzeroca else +fi % if ca is 0, no positive sign before the "x" term if cb is positive
newcommandsignbFPifnegcb -else positiveSignBWithAfi
newcommandsigncFPifnegcc -else +fi
newcommandcoeffaFPifeqaabs1 elseaabsfi
newcommandcoeffbFPifeqbabs1 elsebabsfi
newcommandpolyaFPifzeroca elsesignacoeffa x^2fi
newcommandpolybFPifzerocb elsesignbcoeffb xfi
newcommandpolycFPifzerocc elsesignccabsfi

Quadratic equation : $polya polyb polyc =0$ \[1cm]

Result: $x = xone quad textand quad x = xtwo$

endframe
enddocument