Question

How do you solve `x^2 - 3x - 4 = 0` using the quadratic formula?

Answer 1

`x=-1`
`x=4`

Explanation:

The quadratic formula is `(-b+-sqrt(b^2-4ac))/(2a)`.

In a trinomial written in the form `ax^2+bx+c`,
`a`- the coefficient (number) in front of the `x^2`
`b`- the coefficient in front of the `x`
`c`- the constant (number by itself and has no variable- in this case, has no `x`- attached to it

So, in this problem,
`a= 1` (no coefficient is an invisible `1`)
`b= -3` (don't forget about the negative!)
`c= -4`

Now, we plug all these values into the quadratic formula with the corresponding values:
`(--3+-sqrt((-3)^2-4(1)(-4)))/(2(1))`
`(3+-sqrt(9-4(-4)))/(2)`
`(3+-sqrt(9+16))/(2)`
`(3+-sqrt(25))/(2)`
`(3+-5)/(2)`

Now, you have to solve for BOTH solutions:

One solution: `x=` `(3+5)/2` = `8/2` = `4`

Second solution: `x=` `(3-5)/2` = `-2/2` = `-1`

So, the two solutions are `x= -1` and `x=4`