Решите уравнение х^2-|х+2|=4

$\mathtt{x^2-|x+2|=4~\to~\left[\begin{array}{ccc}\mathtt{x^2+(x+2)=4,~x+2\leq0}\\\mathtt{x^2-(x+2)=4,~x+2\geq0}\end{array}\right}\\$

$\mathtt{\left[\begin{array}{ccc}\mathtt{(x+2)(x-1)=0,~x\leq-2}\\\mathtt{(x+2)(x-3)=0,~x\geq-2}\end{array}\right\left[\begin{array}{ccc}\mathtt{\left\{{{x\leq-2}\atop{\left[\begin{array}{ccc}\mathtt{x=-2}\\\mathtt{x=1}\end{array}\right}}\right}\\\mathtt{\left\{{{x\geq-2}\atop{\left[\begin{array}{ccc}\mathtt{x=-2}\\\mathtt{x=3}\end{array}\right}}\right}\end{array}\right\left[\begin{array}{ccc}\mathtt{x=-2}\\\mathtt{x=3}\end{array}\right}$

ответ: $\mathtt{x=-2;~3}$