Question

How do you solve `x^2 + 4x - 21 = 0`?

Answer 1

We are going to solve it with `Delta`

The equation is of the form `ax²+bx+c=0`

`a=1` `b=4` `c=-21`

`Delta=b²-4ac`
`Delta=4²-4*1*(-21)`
`Delta=16+84`
`Delta=100` ( `=sqrt10`)

`x_1=(-b-sqrtDelta)/(2a)`

`x_1=(-4-10)/2`

`x_1=-14/2`

`x_1=-7`

`x_2=(-b+sqrtDelta)/(2a)`

`x_2=(-4+10)/2`

`x_2=6/2`

`x_2=3`

Answer 2

You can solve this by using the general formula for complete quadratic equations.

`x=(-b+-sqrt(b^2-4ac))/(2a)`
For a complete quadratic equation `ax^2+bx+c=0`.
Let's apply it to your equation, which gives:

`x=(-4+-sqrt(16-4*1*(-21)))/2=(-4+-sqrt(100))/2=(-4+-10)/2=-2+-5`
So, we've got two solutions:
`x=3`
and
`x=-7`

Answer 3

A less direct, but perhaps simpler approach is to recognise that if `x^2+4x-21` has linear factors, then they must look like:

`(x+p)(x-q) = x^2+(p-q)x-(pxxq)` where `p, q > 0`

The reason I say it must look like this with one `+` and one `-` is because the sign of the constant term in the original quadratic is negative.

Then `p-q = 4` and `pxxq = 21`.

What pair of factors of `21` differs by `4`?

There are not many choices of factorizations, basically either `1xx21` or `3xx7`.

`7-3 = 4` and `7xx3 = 21`, so let `p = 7` and `q=3`

`x^2+4x-21 = (x+p)(x-q) = (x+7)(x-3)`

This is zero when `x=-7` or `x=3`