f(x) = 3x2-7x - 6

b^2-4ac = 49-(4*3*-6) >0 

there are two real zeroes for this function

x = -2/3 or x= 3

h(x)= x2 - 4x+4

b^2-4ac = 16-(4*1*4) = 0

There is only one real zero; there is only one real solution

x = 2

g(x) = 15x^2 -10x+9

b^2-4ac = 100-(4*15*9) <0

therefore, the are two imaginary solutions: (no real solutions)

x = (5+i(110)^1/2)/15 or (5-i(110)^1/2)/15

f(x)=-x2+6x+17

b^2-4ac = 36+(4*1*17) >0 

There are two real irrational solutions:

x = -(26)^0.5 + 3 or x = (26)^0.5 + 3