If you have a quadratic equation, ax^2+bx+c, the roots are the numbers that make this equal to zero, which means you need to solve the equation ax^2+bx+c=0. The geometrical interpretation of this is that you find the coordinates of the points that your curve cuts the x-axis (because y=0). In some occasions it is easier to find the roots by factorising. Example: x^2-5x+6=0. We need two numbers that multiply to give 6 (the constant term) and add to give 5 (the coefficient of x). It is always easier to "guess" these numbers from their product. In our case these numbers are -3 and -2, because (-3)x(-2)=6 and -3-2=-5. Then your quadratic equation can be written in the form (x-3)(x-2)=0. A bracket times another bracket is equal to zero. This means that the first bracket is zero, so x-3=0 which means that x=3, or the other bracket is zero, so x-2=0 hence x=2. Hence your roots are x=2 and x=3.

Unfortunately not all the quadratics are factorable. Why? Because in some occasions the roots are irrational numbers or fractions. In that case, we use the quadratic formula. Example x^2+x-1=0. It is quite clear that I cannot find two numbers that multiply out to give -1 and add up to +1. The formula is x=(-b+/-sqrt(b^2-4ac))/2a, so all you have to do is determine the values of a, b and c and substitute in the formula. a is the coefficient of x^2 (the number times by x^2). In our case, a=1. B is the coefficient of x (the number times by x). In our example, b=1, and c is the constant (the number without any x's, and in our case it is c=-1 (NEVER forget the signs).

When we substitute we get x=(-1+/-sqrt(1^2-4x1x(-1)))/2x1=(-1+/-sqrt5)/2

Thus, the first root will be (-1+sqrt5)/2 and the other (-1-sqrt5)/2.

I hope I did help :-)