10.1. Find all roots of the equation $(x-a)(x-b)=(x-c)(x-d)$, given that $a+d=b+c=2015$ and $a \neq c$ (the numbers $a, b, c, d$ are not given).
Answer: 1007.5.

First solution. Expanding the brackets, we get $a b-(a+b) x=$ $=c d-(c+d) x$. Substituting $d=2015-a$ and $b=2015-c$, we obtain $a(2015-c)-(a+2015-c) x=c(2015-a)-(c+2015-a) x$. Simplifying and combining like terms, we get $(a-c) 2015=2(a-c) x$. Considering that $a \neq c$, we get $2 x=2015$, which means $x=1007.5$.

Second solution. Since $c=2015-b$ and $d=2015-a$, the parabolas $y=(x-a)(x-b)$ and $y=(x-c)(x-d)$ are symmetric with respect to the line $x=2015 / 2$. Therefore, they intersect at a point lying on this line. If these parabolas had another common point, then the symmetric point to it would also lie on both parabolas, meaning that our equation would have at least three different roots. But then these parabolas would coincide. This contradicts the condition, as in this case $a=d=2015 / 2$ and $b=c=2015 / 2=a$.

Comment. The answer contains variables $a, b, c$ or $d-$ 0 points.

Shown that $x=1007.5$ is a root of the equation, but not shown that there are no other roots - 4 points.