Find all $f(x)$ if $f(1-x)=f(x)+1-2x$? To find one solution I assumed that $f$ is even and rewrote this as $f(x-1)-f(x)+2x=1.$ By just thinking about a solution, I was able to conclude that $f(x)=x^2$ is a solution. However, I am sure that there are more solutions but I don't know how to find them.
 A: Hint: Let $f(x)=x^2+g(x)$. What properties must $g(x)$ have?
A: Here is a proof which finds all $\;f\;$ for which this equality holds, as the OP (long ago) asked.$%
\require{begingroup}
\begingroup
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
%$  We treat this as a problem to try and simplify the equality, and to try and reduce it to its essentials.
We calculate, for every $\;f \in \mathbb R \to \mathbb R\;$,
$$\calc
   \tag{0}
   \langle \forall x :: f(1-x) = f(x)+1-2x \rangle
\op{\tag{*}\equiv}\hint{arithmetic -- to introduce symmetry}
   \langle \forall x :: f(1-x)-(1-x) = f(x)-x \rangle
\op\equiv\hint{abbreviate $\;g(x) = f(x)-x\;$}
   \langle \forall x :: g(1-x) = g(x) \rangle
\op\equiv\hint{substitute $\;x := \tfrac 1 2 +x\;$ -- to introduce symmetry}
   \langle \forall x :: g(\tfrac 1 2 -x) = g(\tfrac 1 2 +x) \rangle
\op\equiv\hint{abbreviate $\;h(x) = g(x+ \tfrac 1 2)\;$}
   \langle \forall x :: h(-x) = h(x) \rangle
\op\equiv\hint{definition of even}
   h\text{ is even}
\endcalc$$
Therefore, if $\;h\;$ is an even function, then (and only then) $\;f(x) = g(x)+x = h(x-\tfrac 1 2)+x\;$ satisfies $\Ref{0}$.
One example of such an even function is $\;h(x) = ax^2 + c\;$, which gives $\;f(x) = ax^2 + (1-a)x + c + \tfrac a 4\;$, essentially lab bhattacharjee's answer.  And $\;f(x) = \cos(x- \tfrac 1 2) + x\;$ is another valid answer.
Note how the very first step $\Ref{*}$ is really the most creative one, and the others are more or less forced by the desire to find symmetry.
$%
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%$
A: HINT:
As $f(x)-x=f(1-x)-(1-x),$ put $f(x)-x=g(x)$
I'm tempted to add this :
(As IvanLoh has pointed out), if we need $f(x)$ in polynomials 
As $f(x)-f(1-x)=2x-1$ which is $O(x^1), f(x)$ can be at most Quadratic
Let $f(x)=ax^2+bx+c$
$\implies 2x-1=f(x)-f(1-x)=ax^2+bx+c-\{a(1-x)^2+b(1-x)+c\}$
$\implies 2x-1=-(a+b)+2(a+b)x^2$
Equating the constants $a+b=1$
and equating the  coefficients of $x,a+b=1\implies b=1-a$
So, any $f(x)=ax^2+(1-a)x+c$ will satisfy the given condition
