given $2f(x) + f(1-x) = x^2$ find $f(-5)$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

A function $f$ has property that $2f(x)+ f(1-x) = x^2$ for any number of x. What is the number for $f(-5)$?
(A) $\frac {15}4$  (B) $\frac {21}5$  (C) $\frac {25}3$  (D) $\frac {14}3$ (E) $\frac {34}3$

It doesn't give us what $f(x)$ is equal to. 
Tried to do comparison
\begin{align}
2f(-5) + f(1-(-5)) &= (-5)^2 \\
2f(-5) + f(6) &= 25 \quad(1) \\
2f(6) + f(1-(6)) &= (6)^2 \\
2f(6) + f(-5) &= 36 \quad(2) \\
\end{align}
so we have the $f(-5)$ and $f(6)$ in both functions, we can add them together
\begin{align}
(1)+(2) \\
[2f(-5) + f(6)] + [2f(6) + f(-5)] &= 25+36 \\
2f(-5) + f(-5) + f(6) + 2f(6) &= 61 \\
3f(-5) + 3f(6) &= 61 \\
f(6) &= \frac {61}3 - f(-5) \quad(3) \\
\end{align}
replacing $f(6)$, at equation $\quad(3)$ to the first equation, $\quad(1)$
\begin{align}
2f(-5) + [\frac{61}{3} - f(-5)] &= 25 \\
2f(-5) - f(-5) &= 25 - \frac {61}3 \\
f(-5) &= \frac{14}{3} 
\end{align}
EDIT: Well, I am blind, saw + to minus

Hint
If (this was the original post) $$2f(x) - f(1-x) = x^2$$ then $$2f(1-x) - f(x) = (1-x)^2$$ So, multiplying the first by $2$ and adding to the second, we have $$3f(x)=2x^2+(1-x)^2$$ 
Edit
By the way, the answer for $x=-5$ is not in the list of choices and your answer is perfectly correct.
You would get $\frac {34}3$ for $f(-3)$. All other cases correspond to $x$ expressed with radicals. Probably one more typo.
You changed the post (from $-$ to $+$) but the same method applies.
