5x^3-6x^2=2x-3 can be written f(x)=5x^3-6x^2-2x+3. When f(x)=0 the solution x will identify the zeroes of the function f(x). At x=-0.68102 (approx) f(x)=0. When x=0 and therefore f(x)=3, we can identify where the graph of f(x) cuts the vertical axis. There appears to be one zero, or solution for x, at x=-0.68102. This value was found by trial and error method, knowing that there was a solution between -1 and 0, because f(x) goes from -6 to +3 in this range.