#z^2>=4(z+3)# can be simplified to
#z^2>=4z+12# or
#z^2-4z-12>=0# or
#z^2-6z+2z-12>=0# or
#z(z-6)+2(z-6)>=0# or
#(z+2)(z-6)>=0#
AS this gives critical values of #z# as #-2# and #6#,
the sign chart of it is
#color(white)(xxxxxxxxxxx)# #-2# #color(white)(xxxxx# #6#
#z+2##color(white)(xxxxxx)##-##color(white)(xxxxxx)##+##color(white)(xxxxxx)##+#
#z-6##color(white)(xxxxxx)##-##color(white)(xxxxxx)##-##color(white)(xxxxxx)##+#
#z^2-4z-12##color(white)(x)##+##color(white)(xxxxxx)##-##color(white)(xxxxxx)##+#
Hence solution is #z<=-2# or #z>=6#
graph{x^2-4x-12 [-10, 10, -15, 15]}
