Срочно осталось 10 мир !!!!

Решите задачу:

$\sqrt{6+x-x^2}\ \textgreater \ 1-x, \\ \left [ {{\left \{ {{1-x\ \textless \ 0,} \atop {6+x-x^2\geq0;}} \right.} \atop {\left \{ {{1-x\geq0,} \atop {6+x-x^2\ \textgreater \ (1-x)^2;}} \right.}} \right. \left [ {{\left \{ {{-x\ \textless \ -1,} \atop {x^2-x-6\leq0;}} \right.} \atop {\left \{ {{-x\geq-1,} \atop {6+x-x^2\ \textgreater \ 1-2x+x^2;}} \right.}} \right. \left \{ {{\left \{ {{x\ \textgreater \ 1,} \atop {x^2-x-6\leq0;}} \right.} \atop { \left \{ {{x\leq-1,} \atop {2x^2-3x-5\ \textless \ 0;}} \right.}} \right.$
$x^2-x-6=0, \\ x_1=-2, \ x_2=3; \\ 2x^2-3x-5=0, \\ D=49, \\ x_1=-1, x_2=2,5; \\ \left [ {{\left \{ {{x\ \textgreater \ -1,} \atop {(x+2)(x-3) \leq 0;}} \right} \atop {\left \{ {{x\leq-1,} \atop {(x+1)(x-2,5)\ \textless \ 0;}} \right}} \right.\left [ {{\left \{ {{x\ \textgreater \ -1,} \atop {-2\leq x\leq3;}} \right} \atop {\left \{ {{x\leq-1,} \atop {-1\ \textless \ x\ \textless \ 2,5;}} \right}} \right. \left [ {{-1\ \textless \ x\leq3,} \atop {x\in\varnothing;}} \right. \\ -1\ \textless \ x\leq3, \\ x\in(-1;3].$