How do you find the zeros, if any, of #y=x^2-3x+1# using the quadratic formula?
The solutions are
Given a quadratic equation
the quadratic formula states that the solutions, if any, are
#Delta>0#, then the square root is well defined and not null, so you have two solutions.
#Delta=0#, the square root is zero as well. So, adding or subtracting it makes no difference, and you have two coincident solutions.
#Delta < 0#, the square root is not defined, and you have no solutions (at least using real numbers).
In your case,