Question

How do you solve `x^2-2x-1=x+3`?

Answer 1

`x=4`
`x=-1`

Explanation:

First, get `0` on one side.
`(x^2-2x-1)` `-(x+3)`=`x+3` `-(x+3`)

`x^2-3x-4=0`

Next, factor this equation.
`(x-4)(x+1)=0`
Therefore,
`(x-4)=0`
`x=4`
or
`(x+1)=0`,
`x=-1`

You can verify this using FOIL:
=`x^2+x-4x-4`
=`(x^2-3x-4)`

Answer 2

`x=-1`
`x=4`

Explanation:

`x^2-2x-1=x+3`

`x^2-2x-1-x-3=0`

`x^2-3x-4=0`

`x^2-4x+x-4=0`

`x(x-4)+1(x-4)=0`

`(x+1)(x-4)=0`

`x+1=0`
`x=-1`

`x-4=0`
`x=4`

Answer 3

`x=-1,x=4`

Explanation:

Consider the given equation
`x^2-2x-1=x+3`

Subtract `x+3` from both sides (Or bring all terms to the left-hand side)

`x^2-2x-1-x-3=x+3-x-3`

Thus,
`x^2-2x-1-x-3=0`

Add like terms
`x^2-3x-4=0`

Split middle term so as to obtain `(-4)` as product and `(-3)` as sum
`x^2+x-4x-4=0`

Take common factors out from paired terms
`x(x+1)-4(x+1)=0`

Then,
`(x+1)(x-4)=0`
`x+1=0` or `x-4=0`
`x=-1` or `x=4`