Let's rewrite the equation
#2x+2>x^2#
#x^2-2x-2<0#
The roots of the equation
#x^2-2x-2=0#
are #x_1# and #x_2#
#x_2=(2+sqrt(2^2-(4*1*-2)))/2=(2+sqrt12)/2=(2+2sqrt3)/2=1+sqrt3#
#x_1=(2-sqrt(2^2-(4*1*-2)))/2=(2-sqrt12)/2=(2-2sqrt3)/2=1-sqrt3#
Let #f(x)=x^2-2x-2#
We build the sign chart
#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##x_2##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x-x_1##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x-x_2##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
Therefore,
#f(x)<0#, when #x in (x_1,x_2)#