If b^2 -4ac > 0, there are two solutions

If it =0 , there is one solution

if it <0 , there are no solutions

For the second equation above, you must put it into the form ax^2+bx+c=0, so convert it to x^2-2x+1=0. Then use the quadratic formula. We have [2+/- sqrt(4-4)]/2], which is 2/2 = 1. x=1 is the only solution because 2-sqrt(0) and 2+sqrt(0) is the same thing.

For the third equation we have [1+/- sqrt(1-4(3)(-1)]/2(3), or [1-sqrt(13)]/6 and [1+sqrt(13)]/6 — two solutions.

Visit the following site: http://www.purplemath.com/modules/quadform.htm

It’ll give you additional info on the quadratic formula.