Which equations have no real solution but have two complex solutions?

Answer:
2x^2 = 6x - 5 and 3x^2 - 5x = -8
Step-by-step explanation:
Well you can move around the terms in each equation to get it in the form ax^2 + bx + c. From there you can evaluate the value of the discriminant, which is the value under the radical in the quadratic equation. If the discriminant is negative that means you have two complex solutions. The discriminant is equal to b^2 -4ac
3x^2 - 5x = -8
3x^2 - 5x + 8 = 0
discriminant = (-5)^2 - 4(3)(8)
discriminant = 25 - 96
discriminant = -71
2x^2 = 6x - 5
0 = -2x^2 + 6x - 5
6^2 - 4(-2)(-5)
36 - 40
discriminant = -4
12x = 9x^2 + 4
0 = 9x^2 - 12x + 4
(-12)^2 - 4(9)(4)
144 - 144
0 (not negative, just one solution at the vertex)
-x^2 - 10x = 34
-x^2 - 10x + 34 = 0
(-10)^2 - 4(-1)(34)
100 + 136
236