How do you find the roots, real and imaginary, of `y=x(x-1)-812` using the quadratic formula?

`y = x(x-1)-812`

Before we use the quadratic formula, we have to make the equation in standard quadratic form, or `y = ax^2 + bx + c`.

Let's distribute the `x(x-1)`:
`y = x^2 - x - 812`

As you can see, this is now in the form `y = ax^2 + bx + c`, where `color(red)(a = 1)`, `color(magenta)(b = -1)`, and `color(blue)(c = -812)`.

The quadratic equation is used to find the real and imaginary roots/zeros of a quadratic equation. It's formula is `x = (-color(magenta)(b) +- sqrt((color(magenta)(b))^2-4color(red)(a)color(blue)(c)))/(2color(red)(a))`

Let's plug in our values into the formula and solve for `x`:
`x = (-(color(magenta)(-1)) +- sqrt((color(magenta)(-1))^2 - 4(color(red)(1))(color(blue)(-812))))/(2(color(red)(1))`

Now simplify:
`x = (1 +- sqrt(1 + 3248))/2`

`x = (1+- sqrt(3249))/2`

`x = (1+-57)/2`

`x = (1+57)/2`, `x = (1-57)/2`

`x = 58/2`, `x = -56/2`

Therefore, the roots are at:
x = -28, 29

To show that this is correct, I graphed the original equation `y = x(x-1) - 812` using desmos.com:

As you can see, the roots are indeed at `x = -28` and `x = 29`.

If you need another example, feel free to watch this Khan Academy video:

Hope this helps!

`y = x^2 - x - 812`

Discriminant `D = (b^2 - 4ac)`

If D is positive, only real root.

If it’s 0, it’s a perfect square.

If it’s negative, roots imaginary.

In our case, a = 1, b = -1, c = -812.

Hence, D = -1^2 - (4 * 1 * -812) = 3249#

Therefore, both roots are real.

Quadratic roots formula `= (-b +- sqrt (d) ) / (2a)`

`x = (1 +- sqrt 3249) / 2 = (1 +- 57) / 2 = 29, -28`