Question

How do you solve `y^2 - 4y - 21 = 0` using the quadratic formula?

Answer 1

`x_1=7`
`x_2=-3`

Explanation:

So basically the quadratic equation is `x=(-b+-sqrt(b^2-4ac))/(2a)`
Here `x=y, a=1, b=-4, c=-21`
Plug that in and we get:
`x=(-(-4)+-sqrt(4^2-4*1*(-21)))/(2*1)`

`x=(4+-sqrt(16+84))/2`

`x=(4+-sqrt100)/2`

`x=(4+-10)/2`

`x=2+-5`

There are two possible values for `x`.
`x_1=7`
`x_2=-3`