Помогите пожалуйста решить: √(x+2) больше x

[latex] \sqrt{x+2} \ \textgreater \ x[/latex] 1 случай: [latex] \left \{ {{x\ \textless \ 0} \atop {x+2 \geq0 }} \right. [/latex] [latex] \left \{ {{x\ \textless \ 0} \atop {x \geq-2 }} \right. [/latex] x∈ [latex][-2;0)[/latex] 2 случай:  [latex] \left \{ {{x \geq 0} \atop {( \sqrt{x+2})^2\ \textgreater \ x^2 }} \right. [/latex] [latex] \left \{ {{x \geq 0} \atop {x+2\ \textgreater \ x^{2} }} \right. [/latex] [latex] \left \{ {{x \geq 0} \atop {x+2- x^{2} \ \textgreater \ 0} }} \right. [/latex] [latex] \left \{ {{x \geq 0} \atop { x^{2} -x-2\ \textless \ 0} }} \right. [/latex] [latex]D=1+8=9[/latex] [latex]x_1=2[/latex] [latex]x_2=-1[/latex] [latex] \left \{ {{x \geq 0} \atop {(x-2)(x+1)\ \textless \ 0} }} \right. [/latex] x∈ [latex][0;2)[/latex] общее решение :  x ∈ [latex][-2;2)[/latex]