Which equations have no real solution but have two complex solutions? A 3x>2-5=-8
b 2x>2=6x-5
c 12x=9x>2 +4
d -x>2 - 10x =34

see details below

Step-by-step explanation:

First, assume >2 was meant to be &super;. or square.

then the given equations are:

A 3x^2-5=-8

b 2x^2=6x-5

c 12x=9x^2 +4

d -x^2 - 10x =34

In all cases, rearrange equation to identify parameters a, b and c, then calculate determinant using D=b^2-4ac.

A 3x^2x-5=-8

   3x^2-5x+8=0, a=3, b=-5, c=8

   D=b^2-4ac = (-5)^2-4*3*8 = 25 - 96  = -71  < 0   => 2 complex solutions

B 2x^2=6x-5

   2x^2-6x+5, a=2, b=-6, c=5

   D=b^2-4ac = (-6)^2-4*2*5 = 36 - 40 = -4  < 0   => 2 complex solutions

C 12x=9x^2 +4

   9x^2-12x+4, a = 9, b=-12, c=4

   D=b^2-4ac = (12)^2-4*9*4 = 144 - 144 = 0   => 2 real coincident solutions    

D -x^2 - 10x =34

   -x^2-10x-34, a = -1, b=-10, c=-34

    D=b^2-4ac = (-10)^2-4*(-1)*(-34) = 100 - 136 = -36   <0  => 2 complex  solutions    

the last choice

step-by-step explanation:

idk

step-by-step explanation:

idk