How do you find the roots, real and imaginary, of `y=(x/5-5)(-3x-2)` using the quadratic formula?

First, we need to write this equation in standard form:

`y = (color(red)(x/5) - color(red)(5))(color(blue)(-3x) - color(blue)(2))` becomes:

`y = -(color(red)(x/5) xx color(blue)(3x)) - (color(red)(x/5) xx color(blue)(2)) + (color(red)(5) xx color(blue)(3x)) + (color(red)(5) xx color(blue)(2))`

`y = -3/5x^2 - 2/5x + 15x + 10`

`y = -3/5x^2 - 2/5x + (5/5 * 15)x + 10`

`y = -3/5x^2 - 2/5x + 75/5x + 10`

`y = -3/5x^2 + 73/5x + 10`

We can now use the quadratic equation to solve the equation. The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - 4color(red)(a)color(green)(c)))/(2color(red)(a))`

Substituting:

`color(red)(-3/5)` for `color(red)(a)`

`color(blue)(73/5)` for `color(blue)(b)`

`color(green)(10)` for `color(green)(c)` gives:

`x = (-color(blue)(73/5) +- sqrt(color(blue)(73/5)^2 - (4 * color(red)(-3/5) * color(green)(10))))/(2 * color(red)(-3/5))`

`x = (-color(blue)(73/5) +- sqrt(5329/25 - (-120/5)))/(-6/5)`

`x = (-color(blue)(73/5) +- sqrt(5329/25 - (5/5 * -120/5)))/(-6/5)`

`x = (-color(blue)(73/5) +- sqrt(5329/25 - (-600/25)))/(-6/5)`

`x = (-color(blue)(73/5) +- sqrt(5329/25 + 600/25))/(-6/5)`

`x = (-color(blue)(73/5) +- sqrt(5929/25))/(-6/5)`

`x = (-color(blue)(73/5) + 77/5)/(-6/5)` and `x = (-color(blue)(73/5) - 77/5)/(-6/5)`

`x = (4/5)/(-6/5)` and `x = (-150/5)/(-6/5)`

`x = -(4 * 5)/(5 * 6)` and `x = (150 * 5)/(5 * 6)`

`x = -4/6` and `x = (150)/(6)`

`x = -2/3` and `x = 25`