First add #5x-6# to both sides to get:

#6x^2+5x-6 = 0#

This is of the form #ax^2+bx+c = 0#, with #a=6#, #b=5# and #c=-6# so the discriminant is given by the formula:

#Delta = b^2-4ac#

#= 5^2-(4xx6xx-6) = 25 + 144 = 169 = 13^2#

That #Delta# is positive means that the equation has two distinct real solutions. That #Delta# is a perfect square means that the solutions are rational.

The other possible cases are:

#Delta = 0# The equation has one repeated, real, rational root.

#Delta < 0# The equation has no real roots. It has two distinct complex roots (which are complex conjugates).

In our case, the solutions are given by the formula:

#x = (-b+-sqrt(Delta))/(2a) = (-5+-13)/12#

That is #x = -18/12 = -3/2# or #x = 8/12 = 2/3#

This also means that #6x^2+5x-6#

can be factored as #(2x+3)(3x-2)#