How do you solve `x^2+8x+15=0` using the quadratic formula?

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(1)` for `color(red)(a)`

`color(blue)(8)` for `color(blue)(b)`

`color(green)(15)` for `color(green)(c)` gives:

`x = (-color(blue)(8) +- sqrt(color(blue)(8)^2 - (4 * color(red)(1) * color(green)(15))))/(2 * color(red)(1))`

`x = (-color(blue)(8) +- sqrt(64 - 60))/2`

`x = (-color(blue)(8) - sqrt(4))/2` and `x = (-color(blue)(8) + sqrt(4))/2`

`x = (-color(blue)(8) - 2)/2` and `x = (-color(blue)(8) + 2)/2`

`x = (-10)/2` and `x = (-6)/2`

`x = -5` and `x = -3`

The Solution Set Is: `x = {-5, -3}`

Quadratic formula gives the solution of a quadratic equation. For a quaddratic equation `ax^2+bx+c=0`, the roots given by quadratic formula are `(-b+-sqrt(b^2-4ac))/(2a)`.

Hence for `x^2+8x+15=0`, we have `a=1`, `b-8` and `c=15` and hence solution is

`(-8+-sqrt(8^2-4*1*15))/(2*1)`

= `(-8+-sqrt(64-60))/2`

= `(-8+-sqrt4)/2`

= `(-8+-2)/2`

i.e. either `x=-5` or `x=-3`