9. Solve the equation $f(x)=0$, given that the function $f(x)$ is even and for any $x, y$ the equality $f(x+y)=f(x)+f(y)+$ $xy+2014$ holds.
Answer: $\pm 2 \sqrt{1007}$. Solution. Substituting $x=y=0$, we get that $f(0)=$ -2014. Substitute $y=-x$ and use the evenness of $f(x)$, then $-2014=$ $f(0)=2 f(x)-x^{2}+2014$, from which $f(x)=\frac{x^{2}}{2}-2014$. Setting it to zero and solving the obtained equation, we find $x= \pm 2 \sqrt{1007}$.