How do you solve `2x^2 - 4x - 5=0` using the quadratic formula?

`2x^2-4x-5 = 0`

is of the form:

`ax^2+bx+c = 0`

with `a=2`, `b=-4` and `c=-5`.

This has roots given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (4+-sqrt((-4)^2-4(2)(-5)))/(2(2))`

`color(white)(x) = (4+-sqrt(16+40))/4`

`color(white)(x) = (4+-sqrt(56))/4`

`color(white)(x) = (4+-sqrt(2^2*14))/4`

`color(white)(x) = (4+-2sqrt(14))/4`

`color(white)(x) = 1+-sqrt(14)/2`

`color(white)()`
Footnote

The quadratic formula is very useful, but is it just a "magical" formula to you, or do you know how to derive it?

Here's one way:

Given:

`ax^2+bx+c = 0`

We find:

`0 = 1/a(ax^2+bx+c)`

`color(white)(0) = x^2+b/ax+c/a`

`color(white)(0) = x^2+2b/(2a)x+b^2/(2a)^2-b^2/(2a)^2+c/a`

`color(white)(0) = (x+b/(2a))^2-(b^2-4ac)/(4a^2)`

Add `(b^2-4ac)/(4a^2)` to both ends and transpose to get:

`(x+b/(2a))^2 = (b^2-4ac)/(4a^2)`

Take the square root of both sides, allowing for both positive and negative square roots to find:

`x+b/(2a) = +-sqrt((b^2-4ac)/(4a^2))= +-sqrt(b^2-4ac)/(2a)`

Subtract `b/(2a)` from both ends to find:

`x = -b/(2a)+-sqrt(b^2-4ac)/(2a) = (-b+-sqrt(b^2-4ac))/(2a)`