I assume that you are trying to find the roots of "y = x^2 - 10x - 53".

The correct way to solve this one is to "complete the square".

But, on a TI-84 Plus, you can either graph it and see where it crosses the x axis by using the trace function, or you can use the "Apps" button for 9:PlySmlt2 to get to the Polynomial and Simultaneous Equation solver. Use the Polynomial Solver to get approximations of x1=13.83176 and x2=-3.831676.

An "exact" answer can be found by either completing the square, or buying a calculator that has a CAS (Computer Algebra System) built-in, like the TI-89 or one of the other advanced TI models.

To complete the square, start by setting the x expression equal to zero, and moving the constant term to the other side.

x^2 - 10x = 53

Next, take the x coefficient (-10 in this case), divide it by 2 and square the result to get 25. Add the 25 to both sides.

x^2 - 10x + 25 = 53 + 25

At this point you've got a perfect square on the left side. x^2 - 10x + 25 can be replaced by (x - 5)^2. On the right side, just add them together to get 78.

(x - 5)^2 = 78

Subtract 78 from both sides to get:

(x - 5)^2 - 78 = 0

You now have the difference of two squares on the left side (assuming you know the square root of 78) and can re-write as:

((x - 5) + sqrt(78)) ((x - 5) - sqrt(78)) = 0

This means the roots are:

x = 5 - sqrt(78) or x = 5 + sqrt(78)