2.1. Find the sum of all integer values of $a$ from the interval $[-2012 ; 2013]$, for which the equation $(a-3) x^{2}+2(3-a) x+\frac{a-7}{a+2}=0$ has at least one solution.
Answer: 2011. Solution. When $a=3$ there are no solutions. For other $a$, it should be $\frac{D}{4}=(a-3)^{2}-\frac{(a-3)(a-7)}{a+2} \geq 0$. The last inequality (plus the condition $a \neq 3$) has the solution $a \in(-\infty ;-2) \bigcup\{1\} \cup(3 ;+\infty)$. The desired sum is: $-2012-2011-\ldots-5-4-3+1+4+5+\ldots 2013=-3+1+2013=2011$.